Oh yes, the integral sin(θ) won't be zero. I evaluated it correctly during the examination but did it wrong while posting up my attempt.haruspex said:Your first line is wrong. How did you get 1+sin(θ)? You're measuring it down fron horizontal, so should be minus.
And in the last step, you've evaluated the integral of the sin(θ) part to zero. It won't be.
This approach is much easier. I am not much familiar with the instantaneous axis of rotation but this worked out pretty well. Thanks for the alternate method.haruspex said:Another approach is to say the angular speed is v/R and work out the distance from the element to the point of contact.
No, you're measuring theta anticlockwise in the diagram from the "9 o'clock" position, so it will be 0 to pi.Pranav-Arora said:One more question, would the limits for integration change from 0 to -pi?
haruspex said:No, you're measuring theta anticlockwise in the diagram from the "9 o'clock" position, so it will be 0 to pi.