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Rotational acceleration of a post

  • Thread starter C.E
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  • #1
C.E
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1. Hi, I am massively stuck on the following question and am examined on this stuff soon, could someone please help me?

A thin, uniform 15kg post of length 1.75 m is held vertically using a cable attached to the top of the post. A string attached to a 5kg mass passes over a smooth, massles pulley and is attached to the post 0.5m from the top. The post has a pivot at the bottom of it (i.e. that is what it rests on). Suddenly the cable snaps.

a. Find the angular acceleration of the post (with respect to the pivot) just after the cable snaps.
b.Will the angular acceleration from a be constant before the post hits the pulley? Why?
c.What is the acceleration of the 5kg mass the instant the pulley breaks? Is this constant? Why?

b[2].According to the back of my book the answers I should get are:
a. 2.65 rad/s^2
b. no, no explanation why given.
c. 3.31 m/s^2, n (no explanation given)

b[3].My attempt.
a. Torque (T)= moment of inertia (I) x angular acceleration (A)
T= 5g(1.75-0.5)=61.25
I= (MR^2)/4 = 15.3125
a=T/I = 4
(I don't see where I am going wrong for this part, any ideas?)
b. I think no because as the stick falls the line of action of the applied force changes and hence so does the torque. Is this explanation correct?
c. I though this acceleration was just g. As for whether it is constant I thought yes, but the answer is no.
 

Answers and Replies

  • #2
Doc Al
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A thin, uniform 15kg post of length 1.75 m is held vertically using a cable attached to the top of the post. A string attached to a 5kg mass passes over a smooth, massles pulley and is attached to the post 0.5m from the top.
A diagram would help. What's the orientation of the string? What's the pulley attached to?
 
  • #3
Doc Al
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44,877
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b[3].My attempt.
a. Torque (T)= moment of inertia (I) x angular acceleration (A)
T= 5g(1.75-0.5)=61.25
I= (MR^2)/4 = 15.3125
a=T/I = 4
(I don't see where I am going wrong for this part, any ideas?)
Hint: Find the tension in the string once the cable is cut. It's not simply equal to the weight of the 5kg mass.

(Assuming the string is horizontal, I get the book's answer.)
 
  • #4
C.E
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Yes, the string is horizontal. It does not say anything about what the pully is attached to. Why is the tension not just 5g? What am I missing?
 
  • #5
Doc Al
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What am I missing?
The fact that the mass accelerates.
 
  • #6
C.E
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What do you mean?
 
  • #7
Doc Al
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What do you mean?
The tension in the string will equal the weight of the hanging mass when the mass is stationary (not accelerating). (mg - T = 0) But when the mass starts accelerating, the string tension is less. (mg - T = ma)
 

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