Magnitude of Final Angular Speed of Rolling Sphere

[hime]

Homework Statement

A block with mass 2 kg is hanging from a string that goes over a solid disk pulley with mass mdisk = 2 kg and radius Rdisk = 0.3 m with tension T1=11.34N between them. The other end of the string is attached to a massless axel through the center of an sphere on a flat horizontal surface that rolls without slipping and has mass msphere = 3 kg and radius Rs = 0.2m with tension T2=14.04 N. The system is released from rest with acceleration of the system a= 2.7m/s.

What is the magnitude of the final angular speed of the sphere after the mass has fallen the 1 m?

Homework Equations

Let w=omega, and *a be the angular accleration
w^2=wi^2+2(*a)(Theta2-theta1)
w=v/R
*a=a/R

The Attempt at a Solution

v=sqrt(2a(xf-xi))=sqrt(2*2.7*1)=2.32m/s
w=v/Rsphere=2.32/.2=11.169rad/s*s

Did I do it right?

 

[mark726]

Dear student,

Thank you for your post. Your attempt at solving the problem is on the right track, but there are a few things that need to be corrected.

Firstly, your equation for angular acceleration is incorrect. It should be *a = a/R, where a is the linear acceleration and R is the radius of the sphere.

Secondly, your calculation for the final angular speed is incorrect. It should be w = v/Rdisk + v/Rsphere, where v is the linear speed calculated correctly as 2.32 m/s.

Finally, it is important to note that the system is released from rest, so the initial angular velocity is zero. Therefore, the final angular speed will be equal to the calculated value of 11.169 rad/s.

I hope this helps and good luck with your studies!