A hollow spherical shell with mass 2.05 kg rolls without slipping down a slope t

Question

A hollow spherical shell with mass 2.05 kg rolls without slipping down a slope that makes an angle of 40.0 degree with the horizontal. Find the magnitude of the acceleration a_ of the center of mass of the spherical shell. Take the free-ball acceleration to be g = g 9.80 m/s^2. Find the magnitude of the frictional force acting on the spherical shell. Take the free-ball acceleration to be g = g 9.80 m/s^2. Find the minimum coefficient of friction mu needed to prevent the spherical shell from slipping as it rolls down the slope.

Explanation / Answer

Here,

m = 2.05 Kg

theta = 40 degree

part A) for the acceleration a

a = m * g * sin(theta)/(I/r^2 + m)

a = 2.05 * 9.8 * sin(40)/(2/3 * 2.05 + 2.05)

a = 3.78 m/s^2

part B) for the frictional force

2.05 * 9.8 * sin(40) - f = 2.05 * 3.78

solving for f

f = 5.17 N

frictional force is 5.17 N

part c)

minimum coefficient of friction = frictional force/normal force

minimum coefficient of friction = 5.17/(2.05 * 9.8 * cos(40))

minimum coefficient of friction = 0.34

the minimum coefficient of friction is 0.34

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