Answer Torque Req'd to Reach 300 rev/min in 10s - K.E. at 300rev/min

[John O' Meara]

A grindstone in the form of a solid cylinder has a radius of .5m and a mass of 50kg.
(a) What torque will bring it from rest to an angular velocity of 300 rev/min in 10s.
(b) What is its kinetic energy when it is rotating at 300 rev/min.
I will do (b) first: w = 300 rev/min = 5 rev/s
w = 10*pi rad/s
K.E., = .5*I*w^2. Where I = .5*m*R^2...i.e., the moment of inertia. Therefore,
K.E., = 5000*pi^2/16 = 3084J.

(a) P=T*w; T const'. Where P = power, T=torque, w=angular velocity.
(Also work = T*(theta2 - theta1)). But this is only the power for a particular value of w, what is the power when w<300rev/min, e.g., 200, 100, etc., rev/min. We want to find the total amount of K.E., for w=0 to w=300 rev/min interval. Is there an integral some where here, and what is it? Many thanks.
P.S. Currently on the 2nd page of this thread I got a 2nd question on angular velocity, I hope some one will be able to answer it for me. Thanks again.

 

[andrevdh]

b) you made a slight error in calculating the angular velocity (in rev/s)

a) There are equations similar to the constant acceleration equations for the (constant) angular acceleration of an object. In this problem the torque is constant so the angular acceleration of the grindstone will be constant. From the information it should then be possible to calculate the angular acceleration (by using one of the constant angular acceleration equations). Then you need to find the relationship between torque and angular acceleration.