Two objects, M=15.3 kg and m=9.82 kg are connected with an ideal string and susp

Question

Two objects, M=15.3 kg and m=9.82 kg are connected with an ideal string and suspended by a pulley (which rotates with no friction) in the shape of a uniform disk with radius R=7.5 cm and mass Mp=14.1 kg. The string causes the pulley to rotate without slipping. If the masses are started from rest and allowed to move 2.5 m:

(a) What is the final speed of mass m? This part can be done by two methods: either use forces and torques or energy conservation. Please use both the energy method and Newton's second law to double check your answer.

m/s ( ± 0.02 m/s)

(b) What is the final angular speed of the pulley?

rad/s ( ± 0.2 rad/s)

(c) How long did it take for the masses to move from rest to the final position?

s ( ± 0.02 s)

Explanation / Answer

Conservation of Energy :

KE gained by objects M and m and pulley Mp = net PE lost by M and m.
Final KE of m and M = 0.5(M+m)v^2.
Final KE of pulley = 0.5*Iw^2

where I=moment of inertia and w=angular velocity.
For a uniform disk, I = 0.5MpR^2.
Also, v=Rw so
0.5*Iw^2 = 0.5(0.5MpR^2)(v/R)^2 = 0.25Mpv^2.

net PE lost by M and m = (M-m)gh.
Therefore,

0.5(M+m)v^2 + 0.25Mpv^2 = (M-m)gh
v^2 = 4(M-m)gh /[2(M+m) + Mp]

v = sqrt[4(15.3 - 9.82)*9.81*2.5/[2(15.3+9.82) + 14.1]]
v = 2.89 m/sec
(a) Final speed of mass m is v = 2.89 m/s.
(b) Final angular speed of pulley is w = v/R = 2.89/0.075 = 38.53 rad/s.
(c).

The speed of mass m when it has moved distance h is (from the equation for v) :
v = dh/dt = K (h)^0.5

dh/sqrt(h) = K dt
where K=sqrt[4(M-m)g/(2(M+m) + Mp)] =sqrt[4(M-m)g/(2(M+m) + Mp)] = 1.828.

Integrate wrt t :
2sqrt(h) = Kt + C.
When t=0, h=0 so C=0.
Therefore,
the time taken to reach the final position is t = 2sqrt(h)/K

t = 2*sqrt(2.5) / 1.828 = 1.7299 sec

t = 1.73 sec.

Or

(c)

Vf^2 = Vi^2 + 2ay
Vf = Vi + at

since Vi = 0
then a = Vf^2/(2y) = (2.89)^2 / (2(2.5m) = 1.67 m/s^2
t = Vf/a = (2.89m/s)/(1.67 m/s^2) = 1.73 sec

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