 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2016-resit/8/ii/solution

With clockwise rotation taken positive and

v_{0} = b ω_{0}

, the total kinetic energy is

K_{0} = \frac{M}{2} v_{0}^{2} + \frac{I}{2} ω_{0}^{2} = \frac{M}{2} (a^{2} + b^{2}) ω_{0}^{2} .

Angular momentum needs an origin. Its spin about the centre is

I ω_{0}

, in the clockwise direction. About a fixed point in the horizontal rail plane, spin and orbital contributions add to

L_{rail} = I ω_{0} + M b v_{0} = M (a^{2} + b^{2}) ω_{0} .

Here the drum centre is height

b

above that plane. The signed clockwise convention fixes the displayed signs; the associated vector points perpendicular to the vertical plane of motion.

 New to topics? Read the docs here!