Question

a particle starts from rest and moves in a circle with a radius of 1.00 m. its angular position is given by the equation: $\theta = 3.0t - 1.0t^{2}+2.0t^{3}$ where $t$ is in seconds and the angle is in radians. first, determine the instantaneous angular velocity and angular acceleration at 2.00 seconds. second, determine the tangential acceleration and radial acceleration at that time.

Explanation:

Step1: Find the angular - velocity formula

The angular velocity $\omega$ is the derivative of the angular position $\theta$ with respect to time $t$. Given $\theta = 3.0t-1.0t^{2}+2.0t^{3}$, using the power - rule of differentiation $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $\omega=\frac{d\theta}{dt}=3.0 - 2.0t+6.0t^{2}$.

Step2: Calculate the angular velocity at $t = 2.00\ s$

Substitute $t = 2.00\ s$ into the angular - velocity formula: $\omega=3.0-2.0\times2.00 + 6.0\times(2.00)^{2}=3.0-4.0 + 24.0=23.0\ rad/s$.

Step3: Find the angular - acceleration formula

The angular acceleration $\alpha$ is the derivative of the angular velocity $\omega$ with respect to time $t$. Since $\omega=3.0 - 2.0t+6.0t^{2}$, then $\alpha=\frac{d\omega}{dt}=-2.0 + 12.0t$.

Step4: Calculate the angular acceleration at $t = 2.00\ s$

Substitute $t = 2.00\ s$ into the angular - acceleration formula: $\alpha=-2.0+12.0\times2.00=-2.0 + 24.0 = 22.0\ rad/s^{2}$.

Step5: Calculate the tangential acceleration

The tangential acceleration $a_t$ is given by the formula $a_t=r\alpha$, where $r = 1.00\ m$ and $\alpha = 22.0\ rad/s^{2}$. So $a_t=1.00\times22.0=22.0\ m/s^{2}$.

Step6: Calculate the radial acceleration

The radial acceleration $a_r$ is given by the formula $a_r = r\omega^{2}$, where $r = 1.00\ m$ and $\omega = 23.0\ rad/s$. So $a_r=1.00\times(23.0)^{2}=529\ m/s^{2}$.

Answer:

Instantaneous angular velocity at $t = 2.00\ s$: $23.0\ rad/s$
Angular acceleration at $t = 2.00\ s$: $22.0\ rad/s^{2}$
Tangential acceleration at $t = 2.00\ s$: $22.0\ m/s^{2}$
Radial acceleration at $t = 2.00\ s$: $529\ m/s^{2}$