Two objects, M = 15.3 ks and m = 8.29 kg are connected with an ideal string and
ID: 2052388 • Letter: T
Question
Two objects, M = 15.3 ks and m = 8.29 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 = 7.50 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 2.50 m: What is the final speed (m/s) of mass m? What is the final angular speed (rad/s) of the pulley? How long (s) did it take for the masses to move from rest to the final position? Part (a) can be done by two methods: Forces and torques and energy conservation. Please us a second method to check your answer.Explanation / Answer
Part A)
Since the sum of forces acting on each mass must equal ma of the mass, and the weight will be greater than the Tension, we can say that for mass M,
Fw - T1 = Ma
Mg - T1 = Ma
For mass m, the tension is greater than the weight, so
T2 - mg = ma
For the pulley, it has a moment of inertia, and we know that the torque = I
The torque is caused by the force involved in the string tensions on the two sides of the pulley
(T1 - T2)(r) = .5Mpr2 and we know that = a/r
(T1 - T2)(r) = .5Mpr2a/r thus the r cancels out
(T1 - T2) = .5Mpa
From equation 1 we can solve for T1 and we get T1 = Mg - Ma
From equation 2 we can solve for T2 and we get T2 = ma + mg
Substitute that into the Torque Equation
Mg - Ma - ma - mg = .5Mpa
Now substitute
(15.3)(9.8) - (15.3)(a) - (8.29)(a) - (8.29)(9.8) = .5(14.1)(a)
Solve for a
68.7 = 30.64a
a = 2.24 m/s2
Now that we know a, we can solve for the final velocity
vf2 = vo2 +2ad
Since it starts from rest,
vf = [(2)(2.24)(2.5)]
vf = 3.35 m/s
Part B
For angular speed
= v/r
= 3.35 / .075
= 44.7 rad/s
Part C
For this part we can use vf = vo + at
t = vf/a
t = 3.35 / 2.24
t = 1.49 s
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