Question

a)

A large 51.8 kg pottery wheel has a radius of 0.74 m, and can be considered as a solid cylinder rotating about an axis through its center. How much work must be done on the pottery wheel to increase the rate at which it is spinning from 83.0 rpm (rotations per minute) to 139.0 rpm? Give your answer in J.

Hint: The moment of inertia for a solid cylinder rotating about its center with mass M and radius R is given by: I = 1/2 MR²

^b)

A centrifuge rotor has a moment of inertia of 3.127×10^-2 kg m². How much energy, in kJ, is required to bring it from rest to 9.0×10³ rpm (revolutions per minute)?

Answer 1

a)Given,

m = 51.8 kg ; r = 0.74 m ; wi = 83 rpm = 8.7 rad/s ; wf = 139 rpm = 14.6 rad/s

I = 1/2 m R^2

I = 0.5 x 51.8 x 0.74^2 = 14.18 kg-m^2

work done required will be:

W = 1/2 I (wf^2 - 2i^2)

W = 0.5 x 14.18 (14.5^2 - 8.6^2) = 966 J

Hence, W = 966 J

b)Given,

I = 3.127 x 10^-2 kg-m^2 ; w = 9 x 10^3 rpm = 942.5 rad/s

W = 1/2 I (wf^2 - wi^2)

W = 0.5 x 0.03127 x (0 - 942.5^2) = -1.39 x 10^4 J

Hence, W = -1.39 x 10^4 J

(neglect negative sign if only magnitude is needed)