Question

97 km/h on a curve of radius 450 m.

1. a 2.00 - kg stone attached to a rope 4.00 m long is whirled in a circle horizontally on a frictionless surface, completing 5.00 revolutions in 2.00 s. calculate the magnitude of tension in the rope.

Explanation:

Step1: Calculate angular velocity

The stone makes 5.00 revolutions in 2.00 s. One - revolution is $2\pi$ radians. So the angular velocity $\omega$ is given by $\omega=\frac{\theta}{t}$, where $\theta = 5\times2\pi$ radians and $t = 2.00\ s$.
$\omega=\frac{5\times2\pi}{2}=5\pi\ rad/s$

Step2: Identify centripetal force formula

The centripetal force $F_c$ acting on the stone is provided by the tension $T$ in the rope. The centripetal - force formula is $F_c = m\omega^{2}r$, where $m$ is the mass of the stone, $\omega$ is the angular velocity, and $r$ is the radius of the circular path. Here, $m = 2.00\ kg$, $\omega=5\pi\ rad/s$, and $r = 4.00\ m$.

Step3: Calculate the tension

Substitute the values into the formula:
$T=m\omega^{2}r=(2.00\ kg)\times(5\pi\ rad/s)^{2}\times4.00\ m$
$T = 2\times25\pi^{2}\times4$
$T = 200\pi^{2}\ N\approx1974\ N$

Answer:

$1974\ N$