Question

A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the horizontal. The ball rolls without slipping down the incline and at the bottom has a speed of 4.9 m/s. How many revolutions does the ball rotate through as it rolls down the incline

Answer 1

Answer:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!  

The ball rotates 6.78 revolutions.

     

Explanation:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!        

At the bottom the ball has the following angular speed:

$\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s$

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

$sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m$

To find the revolutions we need the time, which can be found using the following equation:                

$v_{f} = v_{0} + at$  

$t = \frac{v_{f} - v_{0}}{a}$ (1)

So first, we need to find the acceleration:

$v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}$    (2)  

By entering equation (2) into (1) we have:

$t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}$

Since it starts from rest (v₀ = 0):  

$t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s$

Finally, we can find the revolutions:  

$\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev$

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!