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:
[tex] \omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s [/tex]
Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:
[tex] sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m [/tex]
To find the revolutions we need the time, which can be found using the following equation:
[tex] v_{f} = v_{0} + at [/tex]
[tex] t = \frac{v_{f} - v_{0}}{a} [/tex] (1)
So first, we need to find the acceleration:
[tex] v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L} [/tex] (2)
By entering equation (2) into (1) we have:
[tex] t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}} [/tex]
Since it starts from rest (v₀ = 0):
[tex] t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s [/tex]
Finally, we can find the revolutions:
[tex] \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 [/tex]
Therefore, the ball rotates 6.78 revolutions.
I hope it helps you!
