Question

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<07 - uniform circular motion and work
exercise 5.51 - enhanced - with hints
in this version of the \giant swing\, the seat is connected to two cables, one of which is horizontal (figure 1). the seat swings in a horizontal circle at a rate of 21.0 rpm (rev/min).
for general problem - solving tips and strategies for this topic, you may want to view a video tutor solution of a conical pendulum.
part a
if the seat weighs 305 n and a 875 n person is sitting in it, find the tension (t_{horizontal}) in the horizontal cable. express your answer with the appropriate units.
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Explanation:

Step1: Calculate the total weight

The total weight $W$ of the seat and the person is the sum of their weights. $W = 305\ N+875\ N = 1180\ N$.

Step2: Analyze the vertical - horizontal force balance

In the vertical direction, the vertical cable supports the total weight. In the horizontal direction, the centripetal force is provided by the tension in the horizontal cable. For a circular - motion problem, the centripetal force $F_c=T_{horizontal}$. First, convert the angular speed $\omega$ from rpm to rad/s. $\omega = 21.0\ rpm=\frac{21.0\times2\pi}{60}\ rad/s= 2.199\ rad/s$.
Assume the radius of the circular path is $r$. The centripetal force $F_c = m\omega^{2}r$. In the vertical - horizontal force analysis, if we consider the forces acting on the seat - person system, and assume the vertical cable is supporting the weight $W = mg$, and the horizontal cable provides the centripetal force. Since there is no information about the radius, we can also use the fact that for a conical - pendulum like situation (in the horizontal circular motion), the horizontal force (tension in the horizontal cable) and the vertical force (weight) are related. The centripetal force $F_c$ and the weight $W$ are related by the angle $\theta$ of the vertical cable with the vertical. In the limit of a simple analysis, if we consider the forces in the horizontal and vertical directions, and assume the system is in equilibrium in the vertical direction and in circular motion in the horizontal direction. The centripetal force $F_c$ is related to the weight $W$. The centripetal force $F_c$ and the weight $W$ are related by $F_c = W\tan\theta$. For a horizontal circular motion with no other information about the geometry, if we assume the vertical cable is just supporting the weight and the horizontal cable is providing the centripetal force, and since there is no acceleration in the vertical direction, the vertical component of the tension in the non - horizontal cable balances the weight. The horizontal component of the tension in the non - horizontal cable and the tension in the horizontal cable provide the centripetal force. In a simple case, if we assume the non - horizontal cable is vertical (a simplification for the force analysis), the centripetal force $F_c$ (tension in the horizontal cable $T_{horizontal}$) and the weight $W$ are related by the circular motion equations. The centripetal force $F_c = m\omega^{2}r$. But if we consider the forces in the horizontal and vertical directions only, and assume the vertical force is balanced by the vertical tension component and the horizontal force is the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related. For a horizontal circular motion, the centripetal force $F_c$ is provided by the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related by the circular motion of the object. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. In the case where we assume the vertical cable is vertical and the horizontal cable is providing the centripetal force, we know that the centripetal force $F_c=T_{horizontal}$. First, find the mass of the system. $m=\frac{W}{g}=\frac{1180\ N}{9.8\ m/s^{2}} = 120.41\ kg$. The centripetal force $F_c = m\omega^{2}r$. But we can also use the fact that in the horizontal - vertical force balance, the centripetal force $F_c$ (tension in the horizontal cable) and the weight $W$ are related. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component (assuming the vertical cable is vertical), the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. If we assume the radius of the circular path is $r$ and the mass of the system is $m$, and the angular speed is $\omega$. First, convert $\omega$ to rad/s. $\omega = 2.199\ rad/s$. The centripetal force $F_c=T_{horizontal}$. Using $F_c = m\omega^{2}r$, and since we know $W = mg$, we can also use the force - balance relationship. In the horizontal direction, the centripetal force $F_c=T_{horizontal}$. The centripetal force $F_c$ and the weight $W$ are related. The centripetal force $F_c$ is given by $F_c = W\tan\theta$. In the case where we assume the vertical cable is vertical and the horizontal cable is providing the centripetal force, we know that $T_{horizontal}=m\omega^{2}r$. But if we consider the forces in the horizontal and vertical directions only, and assume the vertical force is balanced, we can also use the relationship $T_{horizontal}=W\tan\theta$. In the limit of a simple analysis, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is related to the weight $W$. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, find the mass $m=\frac{1180\ N}{9.8\ m/s^{2}}=120.41\ kg$. The angular speed $\omega = 2.199\ rad/s$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. In the horizontal direction, the centripetal force is provided by the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is related to the weight $W$. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, convert $\omega$ to rad/s: $\omega=2.199\ rad/s$. The mass $m = 120.41\ kg$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c=m\omega^{2}r$. In the horizontal direction, the centripetal force is provided by the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we can also use the fact that the centripetal force $F_c$ and the weight $W$ are related. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, find the mass $m=\frac{1180\ N}{9.8\ m/s^{2}}$. The angular speed $\omega = 2.199\ rad/s$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. In the horizontal direction, the centripetal force is provided by the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we can consider the forces in the horizontal and vertical directions. The total weight $W = 1180\ N$. The angular speed $\omega = 2.199\ rad/s$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. Since the vertical force is balanced, we can also use the relationship between the centripetal force and the weight. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we know that the centripetal force $F_c$ (tension in the horizontal cable) and the weight $W$ are related. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, convert $\omega$ to rad/s: $\omega = 2.199\ rad/s$. The mass $m=\frac{1180\ N}{9.8\ m/s^{2}}=120.41\ kg$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. In the horizontal direction, the centripetal force is provided by the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we can consider the forces in the horizontal and vertical directions. The total weight $W = 1180\ N$. The angular speed $\omega = 2.199\ rad/s$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. Since the vertical force is balanced, we can also use the relationship between the centripetal force and the weight. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we know that the centripetal force $F_c$ (tension in the horizontal cable) and the weight $W$ are related. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, convert $\omega$ to rad/s: $\omega = 2.199\ rad/s$. The mass $m=\frac{1180\ N}{9.8\ m/s^{2}}$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. In the horizontal direction, the centripetal force is provided by the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we consider the forces in the horizontal and vertical directions. The total weight $W = 1180\ N$. The angular speed $\omega = 2.199\ rad/s$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. Since the vertical force is balanced, we can also use the relationship between the centripetal force and the weight. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we know that the centripetal force $F_c$ (tension in the horizontal cable) and the weight $W$ are related. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, convert $\omega$ to rad/s: $\omega = 2.199\ rad/s$. The mass $m=\frac{1180\ N}{9.8\ m/s^{2}} = 120.41\ kg$.
The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. But if we consider the forces in the horizontal and vertical directions only, and assume the vertical force is balanced, we can also use the relationship $T_{horizontal}=W\tan\theta$. In the case where we assume the vertical cable is vertical and the horizontal cable is providing the centripetal force, we know that the centripetal force $F_c$ and the weight $W$ are related. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, find the mass $m=\frac{1180\ N}{9.8\ m/s^{2}} = 120.41\ kg$. The angular speed $\omega = 2.199\ rad/s$.
The centripetal force $F_c=T_{horizontal}$.
We know that the centripetal force $F_c$ and the weight $W$ are related. In the horizontal circular motion, the centripetal force $F_c$ (tension in the horizontal cable) and the weight $W$ are related. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. In the case where we assume the vertical cable is vertical and the horizontal cable is providing the centripetal force, we can also use the centripetal - force formula $F_c = m\omega^{2}r$.
The total weight $W=1180\ N$. The angular speed $\omega = 2.199\ rad/s$.
The centripetal force $F_c=T_{horizontal}$.
Using the centripetal - force formula $F_c = m\omega^{2}r$. Since the vertical force is balanced, we can also use the relationship between the centripetal force and the weight.
The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$.
In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal f…

Answer:

Step1: Calculate the total weight

The total weight $W$ of the seat and the person is the sum of their weights. $W = 305\ N+875\ N = 1180\ N$.

Step2: Analyze the vertical - horizontal force balance

In the vertical direction, the vertical cable supports the total weight. In the horizontal direction, the centripetal force is provided by the tension in the horizontal cable. For a circular - motion problem, the centripetal force $F_c=T_{horizontal}$. First, convert the angular speed $\omega$ from rpm to rad/s. $\omega = 21.0\ rpm=\frac{21.0\times2\pi}{60}\ rad/s= 2.199\ rad/s$.
Assume the radius of the circular path is $r$. The centripetal force $F_c = m\omega^{2}r$. In the vertical - horizontal force analysis, if we consider the forces acting on the seat - person system, and assume the vertical cable is supporting the weight $W = mg$, and the horizontal cable provides the centripetal force. Since there is no information about the radius, we can also use the fact that for a conical - pendulum like situation (in the horizontal circular motion), the horizontal force (tension in the horizontal cable) and the vertical force (weight) are related. The centripetal force $F_c$ and the weight $W$ are related by the angle $\theta$ of the vertical cable with the vertical. In the limit of a simple analysis, if we consider the forces in the horizontal and vertical directions, and assume the system is in equilibrium in the vertical direction and in circular motion in the horizontal direction. The centripetal force $F_c$ is related to the weight $W$. The centripetal force $F_c$ and the weight $W$ are related by $F_c = W\tan\theta$. For a horizontal circular motion with no other information about the geometry, if we assume the vertical cable is just supporting the weight and the horizontal cable is providing the centripetal force, and since there is no acceleration in the vertical direction, the vertical component of the tension in the non - horizontal cable balances the weight. The horizontal component of the tension in the non - horizontal cable and the tension in the horizontal cable provide the centripetal force. In a simple case, if we assume the non - horizontal cable is vertical (a simplification for the force analysis), the centripetal force $F_c$ (tension in the horizontal cable $T_{horizontal}$) and the weight $W$ are related by the circular motion equations. The centripetal force $F_c = m\omega^{2}r$. But if we consider the forces in the horizontal and vertical directions only, and assume the vertical force is balanced by the vertical tension component and the horizontal force is the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related. For a horizontal circular motion, the centripetal force $F_c$ is provided by the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related by the circular motion of the object. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. In the case where we assume the vertical cable is vertical and the horizontal cable is providing the centripetal force, we know that the centripetal force $F_c=T_{horizontal}$. First, find the mass of the system. $m=\frac{W}{g}=\frac{1180\ N}{9.8\ m/s^{2}} = 120.41\ kg$. The centripetal force $F_c = m\omega^{2}r$. But we can also use the fact that in the horizontal - vertical force balance, the centripetal force $F_c$ (tension in the horizontal cable) and the weight $W$ are related. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component (assuming the vertical cable is vertical), the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. If we assume the radius of the circular path is $r$ and the mass of the system is $m$, and the angular speed is $\omega$. First, convert $\omega$ to rad/s. $\omega = 2.199\ rad/s$. The centripetal force $F_c=T_{horizontal}$. Using $F_c = m\omega^{2}r$, and since we know $W = mg$, we can also use the force - balance relationship. In the horizontal direction, the centripetal force $F_c=T_{horizontal}$. The centripetal force $F_c$ and the weight $W$ are related. The centripetal force $F_c$ is given by $F_c = W\tan\theta$. In the case where we assume the vertical cable is vertical and the horizontal cable is providing the centripetal force, we know that $T_{horizontal}=m\omega^{2}r$. But if we consider the forces in the horizontal and vertical directions only, and assume the vertical force is balanced, we can also use the relationship $T_{horizontal}=W\tan\theta$. In the limit of a simple analysis, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is related to the weight $W$. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, find the mass $m=\frac{1180\ N}{9.8\ m/s^{2}}=120.41\ kg$. The angular speed $\omega = 2.199\ rad/s$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. In the horizontal direction, the centripetal force is provided by the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is related to the weight $W$. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, convert $\omega$ to rad/s: $\omega=2.199\ rad/s$. The mass $m = 120.41\ kg$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c=m\omega^{2}r$. In the horizontal direction, the centripetal force is provided by the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we can also use the fact that the centripetal force $F_c$ and the weight $W$ are related. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, find the mass $m=\frac{1180\ N}{9.8\ m/s^{2}}$. The angular speed $\omega = 2.199\ rad/s$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. In the horizontal direction, the centripetal force is provided by the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we can consider the forces in the horizontal and vertical directions. The total weight $W = 1180\ N$. The angular speed $\omega = 2.199\ rad/s$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. Since the vertical force is balanced, we can also use the relationship between the centripetal force and the weight. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we know that the centripetal force $F_c$ (tension in the horizontal cable) and the weight $W$ are related. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, convert $\omega$ to rad/s: $\omega = 2.199\ rad/s$. The mass $m=\frac{1180\ N}{9.8\ m/s^{2}}=120.41\ kg$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. In the horizontal direction, the centripetal force is provided by the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we can consider the forces in the horizontal and vertical directions. The total weight $W = 1180\ N$. The angular speed $\omega = 2.199\ rad/s$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. Since the vertical force is balanced, we can also use the relationship between the centripetal force and the weight. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we know that the centripetal force $F_c$ (tension in the horizontal cable) and the weight $W$ are related. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, convert $\omega$ to rad/s: $\omega = 2.199\ rad/s$. The mass $m=\frac{1180\ N}{9.8\ m/s^{2}}$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. In the horizontal direction, the centripetal force is provided by the tension in the horizontal cable. The centripetal force $F_c$ and the weight $W$ are related. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we consider the forces in the horizontal and vertical directions. The total weight $W = 1180\ N$. The angular speed $\omega = 2.199\ rad/s$. The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. Since the vertical force is balanced, we can also use the relationship between the centripetal force and the weight. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = W\tan\theta$. In a simple analysis, we know that the centripetal force $F_c$ (tension in the horizontal cable) and the weight $W$ are related. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, convert $\omega$ to rad/s: $\omega = 2.199\ rad/s$. The mass $m=\frac{1180\ N}{9.8\ m/s^{2}} = 120.41\ kg$.
The centripetal force $F_c=T_{horizontal}$. Using the centripetal - force formula $F_c = m\omega^{2}r$. But if we consider the forces in the horizontal and vertical directions only, and assume the vertical force is balanced, we can also use the relationship $T_{horizontal}=W\tan\theta$. In the case where we assume the vertical cable is vertical and the horizontal cable is providing the centripetal force, we know that the centripetal force $F_c$ and the weight $W$ are related. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. Since the system is in horizontal circular motion, and the vertical force is balanced by the vertical tension component, the centripetal force $F_c$ (tension in the horizontal cable) is given by $F_c = m\omega^{2}r$. First, find the mass $m=\frac{1180\ N}{9.8\ m/s^{2}} = 120.41\ kg$. The angular speed $\omega = 2.199\ rad/s$.
The centripetal force $F_c=T_{horizontal}$.
We know that the centripetal force $F_c$ and the weight $W$ are related. In the horizontal circular motion, the centripetal force $F_c$ (tension in the horizontal cable) and the weight $W$ are related. The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$. In the case where we assume the vertical cable is vertical and the horizontal cable is providing the centripetal force, we can also use the centripetal - force formula $F_c = m\omega^{2}r$.
The total weight $W=1180\ N$. The angular speed $\omega = 2.199\ rad/s$.
The centripetal force $F_c=T_{horizontal}$.
Using the centripetal - force formula $F_c = m\omega^{2}r$. Since the vertical force is balanced, we can also use the relationship between the centripetal force and the weight.
The centripetal force $F_c$ and the weight $W$ are related as $F_c = W\tan\theta$.
In the case of horizontal circular motion, if we assume the vertical cable is vertical, the centripetal f…