Question

Problem #2 A hollow ball of radius 0.5 m and mass 4.5 kg is rolling without slipping on a level surface at a constant speed of 4.0 m/s. The ball rolls up a 40° ramp and eventually stops before rolling back down. (the moment of inertia of a hollow ball of mass M and radius R is MR2) Find: (a) the angular (rotational) speed of the ball (in rad/sec) just before it begins to move up the ramp; (b) the rotational kinetic energy of the ball (in J) right before it begins to move up the ramp; (c) how far along the ramp (in m) it moves before it stops;

Answer 1

Given:

R = 0.5 m

M = 4.5 kg

v = 4 m/s

$\theta = 40^{\circ}$

2 I = PMR 3

a) Let $\omega$ be the angular speed.

As the body is rolling without slipping on a fixed surface, then we can write

$\omega = \frac{v}{R}$

Therefore, the angular speed of the body before it begins to move up the ramp is $\omega = \frac{v}{R} = \frac{4}{0.5} = 8 rad/s$ . [answer]

b) The rotational kinetic energy of the body before it begins to move up the ramp is $K.E_{r}=\frac{1}{2}I\omega^{2}=\frac{1}{2}*\frac{2}{3}*MR^{2}*\omega^{2} = \frac{1}{2}*\frac{2}{3}*4.5*0.5^{2}*8^{2}=24J$ [answer]

c) The translational kinetic energy of the body before it begins to move up the ramp is $K.E_{t}=\frac{1}{2}mv^{2}=\frac{1}{2}*4.5*4^{2}=36J$ .

As there is only work done by gravity, while it moves up the ramp, we can apply the conservation of mechanical energy.

Applying the conservation of mechanical energy:

gain in potential energy = loss in kinetic energy(translational + rotational)

$\Rightarrow mgh = K.E_{t}+K.E_{r}$ [where h = height reached by the body]

$\Rightarrow h = \frac{K.E_{t}+K.E_{r}}{mg}$

$\Rightarrow h = \frac{36+24}{4.5*9.81}m$

Therefore, the distance travelled along the ramp is $s = \frac{h}{sin\theta} =\frac{\frac{36+24}{4.5*9.81}}{sin40^{\circ}}=\frac{60}{4.5*9.81*sin40^{\circ}}=2.11m$ [answer]