Question

Two identical bowling balls are rolling on a horizontal floor without slipping. The initial speed of both balls is v = 10 m/s. Ball A encounters a frictionless ramp, reaching a maximum vertical height HA above the floor. Ball B on the other hand rolls up a regular ramp (i.e. without slipping), reaching a maximum vertical height HB above the floor. Which ball goes higher and by how much? Show the steps of your calculations.

Answer 1

Answer:

The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.

We will use conversation of energy.

$K_A_1 + U_A_1 = K_A_2 + U_A_2\\\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_A$

Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.

For the ball B;

$K_B_1 + U_B_1 = K_B_2 + U_B_2$

$\frac{1}{2}I_B\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_B$

The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.

Hence, both balls climb the same point.

Explanation:

Answer 2

Both balls A and B will climb up to the same height according to the conservation of energy.

Calculating the height:

Since the balls are identical, let both have mass m and velocity v.

Also the moment of inertia I and the angular speed ω will be the same.

Ball A encounters a frictionless ramp:

On a frictionless ramp, the ball slides down the ramp since it cannot roll as there is no friction present. Since there is o frictional force, there is no dissipation of energy. Energy is conserved.

According to the law of conservation of energy, the total energy of the system must be conserved.

KE(initial) + PE(initial) = KE(final) + PE(final)

$\frac{1}{2}I\omega^2+\frac{1}{2}mv^2+0=0+mgH_A\\\\H_A=\frac{1}{mg} [\frac{1}{2}I\omega^2+\frac{1}{2}mv^2]$

initial KE has rotational and translational kinetic energy and the initial PE is zero since the ball is on the ground, also the final KE is zero since the velocity at the highest point will be zero.

Ball B encounters a regular ramp:

On a regular ramp, the ball can roll without sliding due to friction.

$\frac{1}{2}I\omega^2+\frac{1}{2}mv^2+0=0+mgH_B\\\\H_B=\frac{1}{mg} [\frac{1}{2}I\omega^2+\frac{1}{2}mv^2]$

The height will be the same since the velocity is the same as ball A.

Also, the translational or rotational velocity will be zero at the highest point.

Hence, both balls climb the same height.

Learn more about conservation of energy:

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