EXAMPLE 8.12A Ball Rolling Down an Incline A ball starts from rest at the top of

Question

EXAMPLE 8.12A Ball Rolling Down an Incline

A ball starts from rest at the top of an incline and rolls to the bottom without slipping.

GOAL Combine gravitational, translational, and rotational energy.

PROBLEM A ball of mass M and radius R starts from rest at a height of 2.00 m and rolls down a 30.0° slope, as in the figure. What is the linear speed of the ball when it leaves the incline? Assume that the ball rolls without slipping.

STRATEGY The two points of interest are the top and bottom of the incline, with the bottom acting as the zero point of gravitational potential energy. As the ball rolls down the ramp, gravitational potential energy is converted into both translational and rotational kinetic energy without dissipation, so conservation of mechanical energy can be applied.

SOLUTION

Apply conservation of energy withPE = PEg, the potential energy associated with gravity.

(KEt + KEr + PEg)i = (KEt + KEr + PEg)f

Substitute the appropriate general expressions, noting that(KEt)i = (KEr)i = 0 and (PEg)f = 0.

0 + 0 + Mgh =

Mv2 +

MR2

2 + 0

The ball rolls without slipping, soR = v, the "no-slip condition," can be applied.

Mgh =

Mv2 +

Mv2 =

Mv2

Solve for v, noting that M cancels.

LEARN MORE

QUESTION Rank from fastest to slowest: (1) a solid ball rolling down a ramp without slipping, (2) a cylinder rolling down the same ramp without slipping, (3) a block sliding down a frictionless ramp with the same height and slope. (Select all that apply.)

What variables in the problem determine the change in gravitational energy, and the total kinetic energy at each location? Consider how the moment of inertia is changed if the mass is concentrated farther from the axis. How does the moment of inertia for a given radius determine the fraction of the energy from changing height that goes into rotational kinetic energy? How does that affects the remaining kinetic energy and speed at each height?

PRACTICE IT

EXERCISEHINTS:  GETTING STARTED  |  I'M STUCK!

Explanation / Answer

1. Block > ball > cylinder

using conservation of energy for Block

mgh = (0.5) m V2

V = sqrt (2gh)                      for Block

using conservation of energy for ball

mgh = (0.5) m V2 + (0.5) I w2

mgh = (0.5) m V2 + (0.5) (2/5 m r2) (V/r)2

mgh = (0.5) m V2 + (0.5) (2/5 m) V2

gh = (0.5) (7/5) V2

V = sqrt (1.43 gh)                         For ball

using conservation of energy for cylinder

mgh = (0.5) m V2 + (0.5) I w2

mgh = (0.5) m V2 + (0.5) (0.5 m r2) (V/r)2

mgh = (0.5) m V2 + (0.5) (2/5 m) V2

gh = (0.5) (1.5) V2

V = sqrt (1.33 gh)                         For cylinder

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