A massless string is wound around a solid sphere (I=2/5 MR^2, M= 5.00kg, R=0.367
ID: 1437607 • Letter: A
Question
A massless string is wound around a solid sphere (I=2/5 MR^2, M= 5.00kg, R=0.367m). The string is draped across a pulley (I=0.5 MR^2, M=1.50 kg, R=0.30 m) and connected to a 7.00kg mass that is at rest (initially). Assume the pulley and sphere are on frictionless bearings. The mass falls 1.45 m. Answert the questions below.
a) What is the constant acceleration of mass?
b) What is the constant angular acceleration of the sphere?
c) After the mass has fallen 1.45 m (and the string has been unwinding without slipping), how fast is it moving? (Keep in mind that there are three strings with the same mass here, so you can't just cancel the masses like they are all the same).
Explanation / Answer
on mass : (using Fnet = ma)
7g - T1 = 7a ........(i)
on pulley:
torque = I x alpha
r ( T1 - T2) = (0.5 M r^2) (a/r)
T1 - T2 = 1.50a/2 = 0.75a ........(ii)
on sphere:
r T2 = (2 MR^2/5)(a/r)
T2 = (5 x 2/5)a = 2a ......(iii)
(i) + (ii) + (iii) ,
7g = (7 + 0.75 + 2)a
a = 7.04 m/s^2
b) angular acceleration, alpha = a/r = 7.04/ 0.367 = 19.17 rad/s^2
c) using vf^2 - vi^2 = 2ad
vf^2 - 0 ^2 = 2(7.04)(1.45)
vf =4.52 m/s
d) w = v/r = 4.52 / 0.367 = 12.31 rad/s
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