Question

A uniform hollow spherical ball of mass 1.75 kg and radius 40.0 cm rolls without slipping up a ramp that rises at 30.0° above the horizontal. The speed of the ball at the base of the ramp is 2.63 m/s. While the ball is moving up the ramp, find (a) the acceleration (magnitude and direction) of its center of mass and (b) the friction force (magnitude and direction) acting on it due to the surface of the ramp.

Answer 1

2.94 m/s²

3.43 N

Step-by-step explanation:

To start with, the combination of rotational and translational kinetic energy can be converted to gravity potential energy. So, we will apply this in solving our question.

PE = KE

mgh = ½mv² + ½Iω²

Moment of inertia of a hollow sphere

I = ⅔mr²

We also know that,

ω = v/r, so if we substitute for ω and I, we have

mgh = ½mv² + ½(⅔mr²)(v/r)², simplifying

gh = ½v² + ½(⅔r²)(v/r)², simplifying further

gh = ½v² + ⅓v², adding up

gh = (5/6)v², making h subject of formula

h = 5v²/6g

Now, we substitute for v and g to have

h = 5(2.63)² / 6(9.81)

h = 34.585 / 58.86

h = 0.588 m

To get the distance that the ball rolls on the ramp before stopping, we say

d = 0.588 / sin30.0

d = 0.588 / 0.5

d = 1.176 m

so acceleration is

a = (2.63² - 0²) / 2(1.176)

a = 6.917 - 0 / 2.352

a = 2.94 m/s

B

Neglecting friction, the sphere acts like a point mass and tends to rise.

mgh = ½mv², simplifying

h = v²/2g

h = 2.63² / 2(9.81)

h = 6.917 / 19.62

h = 0.353 m

The work done by the friction force is the same as the rotational kinetic energy of the sphere, so we say

U = KEr

Fd = ½Iω²

Fd = ½(⅔mr²)(v/r)²

Fd = ⅓mv²

F = mv²/3d

where d is the actual distance up the slope the sphere did roll.

The average friction force is thus

F = 1.75(2.63)² / 3(1.176)

F = 1.75 * 6.917 / 3.528

F = 12.1 / 3.528

F = 3.43 N