Calculating Frictional Force on a Rolling Sphere: Incline Physics Problem

[DragonZero]

Homework Statement

A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle θ = 24°. The sphere has mass M = 8.0 kg and radius R = 0.19 m. The coefficient of static friction between the sphere and the plane is μ = 0.64. What is the magnitude of the frictional force on the sphere?

Homework Equations

So far I can think of these:

torque = inertia * angular acceleration

Force (in direction of ball rolling down) - Friction Force = mass * acceleration, though in this case it's
Force - Friction Force = mass * gravity * sin(angle)

The Attempt at a Solution

Force = 8 * 9.8 * sin(24) = 31.88
inertia = (2/5) * mass * radius^2 = (2/5) * 8 * 0.19^2 = 0.11552

 

[ideasrule]

Force (in direction of ball rolling down) - Friction Force = mass * acceleration, though in this case it's
Force - Friction Force = mass * gravity * sin(angle)

No, the force is just gravity, mg*sin(a). Newton's second law would then be mg*sin(a)-f=ma, where f is friction.

That's the first equation. The second equation you need should be the rotational Newton's second law, expressing angular acceleration in terms of the mass, friction force, and moment of inertia (which you calculated correctly). Consider the contact point between the sphere and the ramp as your reference point for calculating torque.

The final equation you need should relate angular acceleration to linear acceleration. This isn't too hard: it's just alpha=a/R, where R is the radius of the sphere.

 

[DragonZero]

How should i go about calculating torque? I know it's inertia * alpha, but since I don't yet have a linear acceleration I can't use alpha = a/R. I know torque also equals radius * force, so for that force would I plug in m*g*sin(angle)?

 

[ideasrule]

Draw a free-body diagram on the sphere and label ALL forces, not just gravity. You can calculate the torque contribution from each force by multiplying the force by the length of the moment arm. (This length is the perpendicular "distance" between the force vector and the reference point.)

One more thing: the moment of inertia you calculated is about the sphere's center of mass. Try using the parallel axis theorem to calculate the moment of inertia about the contact point.

 

[DragonZero]

torque = r * f = r * m*g*sin(angle) = 0.19 * 8 * 9.8 * sin(24) = 6.05

inertia about contact point = inertia center of mass + mass * distance^2 = 0.11552 + 8 * 0.19^2 = 0.40432

 

[ideasrule]

torque = r * f = r * m*g*sin(angle) = 0.19 * 8 * 9.8 * sin(24) = 6.05

Yes, because the torque contribution from friction and the normal force are both zero.

inertia about contact point = inertia center of mass + mass * distance^2 = 0.11552 + 8 * 0.19^2 = 0.40432

Yes. Now, torque=I*alpha and alpha=a/r, so you can solve the system of equations.

 

[DragonZero]

Thank you.