Hi There! I am at a loss. For #4, how do you get a positive value for the angular speed equation? (part b) When I assumed the torque due to Friction was causing a negative torque, I got a negative alpha, therefore my angular speed equation ended up being w=-3(mu)gt/2R. How do I know to make that torque positive based on the problem? Thank you for your help!

Michelle, when the sphere enters the friction area, it starts rolling in the clockwise direction due to the friction force acting opposite the linear motion (to the left, or negative direction). So, the angular velocity increases with a constant angular acceleration due to the friction force.
Based on that, the angular acceleration is positive, and the equation for angular velocity is positive. The convention that a clockwise rotation is a negative torque and acceleration is only for an isolated rotation - once you have a problem with friction, you're not bound by the convention - just like for falling objects, you can choose down as positive and acceleration as positive.
What's tricky about rolling and sliding is that the linear velocity is decreasing but the angular velocity is increasing. Then, in the last part of the problem both the linear and rotational velocity decrease as it comes to a stop.
John

For #2, I am getting a=2Mg/(4m+m). I am also confused how the cylinder is rolling without slipping if there is no friction (at least there is no mu indicated in the problem or solutions). I have T1=Ma+1/2ma by summing the torques acting on the pulley and T2=Ma by summing the forces in the x-direction on the cylinder.

Daniel, you are correct to assume that there is friction, but fortunately, it cancels out. For the cylinder, T2 is pulling on the axle at the center of the cylinder, so it does not exist a torque, and the equation is T2 - f = Ma. Now the friction is exerting a torque at the bottom of the cylinder. It is a positive torque as it is accelerating the cylinder as the mass falls down. Thus, fR = I alpha. solve those two equations simultaneously (and alfa = a/R since it is rotating without slipping, and you get T2 = 3/2 Ma. Then use Mg - T1 = Ma for the block and T1R - T2R = I alfa, and hopefully it works out for you! If not, please let me know and I'll email you the worked out solution. Thanks for your question. John

## 4 Comments

Michelle Beach •2 years, 4 months ago • login to replyJohn Ennis •2 years, 4 months ago • login to replyDaniel Garza •2 years, 3 months ago • login to replyJohn Ennis •2 years, 3 months ago • login to reply