Rotational Dynamics Questions

  • Thread starter Gimp
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  • #1
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I have a couple questions could someone help? I tried everything and I just don't understand what's going on. Thank you sooo.... much! :D

1)A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of 0.0600 m, an angular speed of 88.0 rad/s, and a moment of inertia of 0.850 kgm2. A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of 5.40 s.
(a) Find the magnitude of the angular deceleration of the cylinder.
(b) Find the magnitude of the force of friction applied by the brake shoe.

2)The crane shown in the drawing is lifting a 190 kg crate upward with an acceleration of 1.8 m/s2. The cable from the crate passes over a solid cylindrical pulley at the top of the boom. The pulley has a mass of 130 kg. The cable is then wound onto a hollow cylindrical drum that is mounted on the deck of the crane. The mass of the drum is 150 kg, and its radius is 0.76 m. The engine applies a counterclockwise torque to the drum in order to wind up the cable. What is the magnitude of this torque? Ignore the mass of the cable.
 

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  • #2
dx
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What have you done?
 
  • #3
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Mainly it's the fact that I really don't understand this stuff at all. I tried different ways to look at it, but all of them don't make sense. I've been staring at this for hours now and still i can't get anything that makes sense, it's really frustrating. I know you guys don't show us how to do it if we don't have any work, but could you just guide me a certain way? I would really appreciate it.
 
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  • #4
dx
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Ok, lets begin with the first one. The average angular acceleration is the change in angular velocity divided by the time taken for that change. Once you get the average acceleration, you can find out the magnitude of the torque from the formula [tex] \tau = I\alpha [/tex]. And then, you can determine the force of friction applied from the formula [tex] r\times{F} = \tau [/tex]. I'll let you do the second one on your own.
 
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  • #5
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Yep, so i took 88.0 rad/s and divded that by 5.4sec and got a angular acceleration of 16.3. Then took that times the I, and that equaled 13.85 for the torque. Now what formula do I use? Is it...

(Force * radius = I * angular acceleration) to find the Force applied to the brake?

and then how about angular decceleration is that any different? or is it just a negative sign in front of the 16.3 that i found before??
 
  • #6
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I got it! Thanks for your help! But can u guide me in a direction with the second one? PLEASE!
 
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