Lab 8: Rotational Dynamics and Moment of Inertia Only 4 more to go!! Rotational motion and linear motion are very similar. Many of the quantities we discuss in rotation have linear motion counterparts: Linear MotionRotational Counterpart Position: s, dAngular position:  Displacement:  x = x F - x i Angular displacement:   =  F -  i Velocity: v =  x /  tAngular velocity:  =  /  t Acceleration: a =  v/  tAngular acceleration:  =   /  t Mass: mMoment of inertia: I ForceTorque Newton's 2 nd law: F = maNewtons’s 2 nd law:  = I 

We can relate these rotational quantities to those analogues linear ones: Linear or tangential velocity:v = r  Tangential or linear acceleration:a = r  Torque: This is the rotational analogue to a linear force. Torque causes rotations. Torque is related to force by the following equation:  = F r sin   r  = F r sin  sin  = d/r so then torque becomes:  = F d We refer to d as the lever arm. F d What if  = 90 o ? The torque becomes:  = F r

Moment of Inertia is the rotational analogue of mass, m I (point mass) = Mr 2 I (ring) = Mr 2 I (disk) = ½ Mr 2 What is the moment of inertia for the “dumbells” shown below? MM r r I (point mass) = Mr 2 Moments of inertia add, so our total I for the two masses is: I = I 1 + I 2 = Mr 2 + Mr 2 = 2Mr 2 What is the effect on I if a triple r? r  3r I = I 1 + I 2 = M(3r) 2 + M(3r) 2 = 9Mr 2 + 9Mr 2 = 18 Mr 2 We increased I by a factor of 9!!

m r How do we find the moment of inertia for the disk in this situation? T mg After drawing the FBD, we can write Newton’s 2 nd Law for the block: Next we write Newton’s 2 nd law for the rotating disk, and since the disk is rotating we need to use the rotational form of Newton’s 2 nd law: Which force is causing the disk to raotate? It’s the force Associated with the tension, T in the string so we can write:

If we use the equation for the block we can solve for tension, T: Substitute this equation back into : And we get: Now solve for I: If instead we solve for 1/a from this equation, we get: