Question

A solid, uniform sphere of mass 2.0 kg and radius 1.7 m rolls from rest without slipping down an inclined plane of height 7.0 m. what is the angular velocity of the sphere at the bottom of the inclined plane?

Answer 1

First of all, to work out the velocity at the bottom you can use the conservation of energy principle as friction isn't specified. So U = T or mgh = $\frac{1}{2} mv^{2}$, rearranging to get $v = \sqrt{2gh}$. From this you get v = 11.72$\frac{m}{s^{2} }$. Following angular velocity ω = $\frac{v}{r}$ which you can use to get ω = 6.9$\frac{rad}{s}$. 

Hope that helps

Answer 2

The angular velocity of the sphere at the bottom of the inclined plane is mathematically given as

w=6.9rad/s

What is the angular velocity of the sphere at the bottom of the inclined plane?

Question Parameter(s):

A solid, uniform sphere of mass 2.0 kg and radius 1.7 m

Rolls from rest without slipping down an inclined plane of height 7.0 m.

Generally, the equation for Torque  is mathematically given as

mgh = 0.5 mv^{2}

Therefore

v =  √{2gh}

v = 11.72

In conclusion,  angular velocity

w =v/r

w=11.7/1.7

w=6.9rad/s

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