A solid sphere of uniform density starts from rest and rolls without slipping do

Question

A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle ? = 30o. The sphere has mass M = 8 kg and radius R = 0.19 m . The coefficient of static friction between the sphere and the plane is ? = 0.64. What is the magnitude of the frictional force on the sphere?

A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle ? = 30 degree . The sphere has mass M = 8 kg and radius R = 0.19 m . The coefficient of static friction between the sphere and the plane is ? = 0.64. What is the magnitude of the frictional force on the sphere?

Explanation / Answer

friction exerts a torque on the sphere; the magnitude of this torque around the centre of the sphereis
f R where f is the force of friction and R is the radius of the sphere

torques cause angular accelerations according to:

torque = I alpha, where I is the moment of inertia and alpha is the angular acceleration

I for a sphere = 2/3 mR^2

alpha = a/R where a is thelinear acceleration. so we have

torque = f R = 2/5mR^2(a/R) or
f= 2/5 ma

now, we know that newton's second law tells us that the sum of forces on the sphere =
mass x linear acceleration of the sphere

component of gravity - friction = ma

component of gravity = mg sin(theta) and friction we know is 2/5 ma

so we have

mgsin(theta) - 2/5ma = ma or a=5/7 g sin(theta)

since friction = 2/5 ma, friction = 2/5(5/7) m g sin(theta) = 2/7 mgsin(theta)

substitute m=8Kg, g=9.8m/s/s and theta =30 deg to find the force of friction on the sphere

Friction = (2/7)*8*9.8*Sin30 = 11.2N

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