Question

A solid sphere with a radius of 5cm initially at rest rolls 270cm down a slope 37° above the horizontal. What is its velocity (rad/s) at that point? [PS: it has not reached the foot of the slope]​

Answer 1

This question involves the concepts of the law of conservation of energy, kinetic energy, potential energy, and moment of inertia.

The angular velocity of the solid sphere at this point is "95.44 rad/s".

According to the law of conservation of energy:

Loss in Potential Energy = Gain in Rotational Kinetic Energy + Gain in Translational Kinetic Energy

$mgh = \frac{1}{2}I\omega^2+\frac{1}{2}mv^2$

where,

m = mass of sphere

g = acceleration due to gravity = 9.81 m/s²

h = loss in height = (270 cm) Sin 37° = 162.5 cm = 1.625 m

r = radius of sphere = 5 cm = 0.05 m

ω = angular speed of sphere = ?

v = linear speed of sphere = rω

I = moment of inertia of sphere = $\frac{2}{5}mr^2$

Therefore,

$mgh=\frac{1}{2}\frac{2}{5}mr^2\omega+\frac{1}{2}m(r\omega)^2\\\\gh=\frac{1}{5}r^2\omega^2+\frac{1}{2}r^2\omega^2\\\\(9.81\ m/s^2)(1.625\ m)=(\frac{1}{5}+\frac{1}{2})(0.05\ m)^2\omega^2\\\\\omega=\sqrt{\frac{15.941\ m^2/s^2}{0.00175\ m^2}}$

ω = 95.44 rad/s

Learn more about the law of conservation of energy here:

brainly.com/question/20971995?referrer=searchResults

The attached picture explains the law of conservation of energy.