Turntable rotation homework

  • Thread starter Gtseviper
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  • #1
Please help

When a turntable rotating at 32 1/3 rev/min is shut off, it comes to rest in 29 s.

a assuming constant angular accelerationn, find the angular acceleration.
(my work) W-Wo/t=a and I got -.12

b number of revolutions
(my work) phida-initial phida= Wot + 1/2at^2 and I got 8.03 rev

2 A solid ball of mass 1.6kg and diameter 20 cm is rotating about its diameter at 64rev/min

a What is its KE.
(my work) K= 1/2 Iw^2, I=2/5 MR^2 I= 2.56 x10^-2 kg m^2 K= 5.75 x 10^-1 J

b addition 2 J is added, new angular speed.
(my work) 2.57 J is new K, W^2= 2K/I W= 1.36 x 10^2 rev/min

3 A 1300-kg car is being unloaded by a winch. At the moment shown in Figure 9-46, the gearbox shaft of the winch breaks, and the car falls from rest. During the car's fall, there is no slipping between the (massless) rope, the pulley, and the winch drum. The moment of inertia of the winch drum is 300 kg · m2 and that of the pulley is 2 kg · m2. The radius of the winch drum is 0.80 m and that of the pulley is 0.30 m. Find the speed of the car as it hits the water.

and I got 8.89 by using mgh= 1/2 Iw (V/rw)^2+ 1/2 Ip (V/rp)^2 mv^2
 

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Answers and Replies

  • #2
malawi_glenn
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For question 1, the turntable. 32 1/3 revs per minute gives ω0 = 0.5389 s-1.
The angular acceleration (assuming constant) gives α = (0-ω0)/t = - 0.5389 / 29 = 0.018582 s.
Can you find the correct number of revolutions now? Keep in mind that one revolution is 2π
 

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