Angular acceleration of solid sphere on frictionless yoke unde no slip roll condition

jvani

1. The problem statement, all variables and given/known data

m=6kg
r=0.18m
Fapp=33N


I'm struggling to understand how to answer this question and correlate the linear force applied to rotation without being given the coefficient of friction causing the rotational motion.

The question states that a solid sphere of 6kg is free to roll on a horizontal surface under no slip conditions. A frictionless axle is run through the middle of the sphere, with a rope thread through it, the rope applies 33N on the sphere, and the sphere starts to rotate without slipping

Any help would be appreciated

Doc Al

Start by identifying the forces acting on the sphere. Then apply Newton's 2nd law to both rotation and translation.

jvani

The only forces I can see in the horizontal direction is the force applied of 33N and the force of friction. I'm struggling see how to find the acceleration of the center of the sphere from what was given, and then how the translational acceleration of the sphere correlates to the rotational acceleration explicitly.

Doc Al

Right.
Apply Newton's 2nd law, as I suggested.
To relate translational and rotational motion, apply the condition for rolling without slipping.

canadiankid

I have the same question, with different numbers. I do not understand which formula I am to use that relates the force applied to both the translational and rotational dynamics. I realize that the tangential acceleration and centre of mass acceleration are the same.

If I say that Fext=ma and then divide by the mass to get the acceleration, and then use the relationship between linear and angular acceleration it does not work.

How do I find the correct acceleration of the centre of mass?

Doc Al

Make sure you use ƩF = ma. To solve for the acceleration, you'll need to combine two equations: one for translation and one for rotation.