Disk and Block

  • Thread starter Awwnutz
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  • #1
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http://img267.imageshack.us/img267/9499/diskwithblockpu2.gif [Broken]

A uniform disk of mass Mdisk = 4.9 kg and radius R = 0.26 m has a small block of mass mblock = 2.6 kg on its rim. The disk rotates about an axis a distance d = 0.15 m from its center that intersects the disk along the radius on which the block is situated.

a) What is the moment of inertia of the block about the rotation axis?

b) What is the moment of inertia of the disk about the rotation axis?

c) When the system is rotating about the axis with an angular velocity of 4.2 rad/s, what is its energy?

d) If while the system is rotating with angular velocity 4.2 rad/s it has an angular acceleration of 8.9 rad/s2, what is the magnitude of the acceleration of the block?




Rotational kinematics



My attempt at the solution is just a bunch of chicken scratch. I had no problem with linear kinematics and my teacher keeps telling us that this is no different between that and rotational kinematics, but for some reason I'm just getting confused. I know: position = theta, velocity = omega, acceleration = alpha, mass = I (center of mass), Force = torque.

I'm not exactly sure how to find the center of mass and then use newton's second law to find alpha and omega. I might be making this harder than it really is.
 
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Answers and Replies

  • #2
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part a for I, i did:

Mblock*distance^2

2.6kg*(.26m-.15m)^2 = .03146kg*m^2, and this is correct.
 
  • #3
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for part b its is telling me to use the parallel axis theorem. I'm not too sure how to set it up. I know its the integral of a distance multiplied by the change in mass over a certain limit. But i don't know where to start.
 
  • #4
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The parallel axis theorem says:

I = Icom + Mass*Radius^2

So i have the mass of the disk, and the radius, how do i find the Icom?
 
  • #5
Icom=1/2MR^2
 

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