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Falling bucket

  • Thread starter MAPgirl23
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  • #1
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A bucket of water of mass 14.9 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.250 m with mass 11.4 kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 10.2 m to the water. You can ignore the weight of the rope.

What is the tension in the rope while the bucket is falling?
With what speed does the bucket strike the water?
What is the time of fall?
While the bucket is falling, what is the force exerted on the cylinder by the axle?

** The gravitational force on the dropping bucket provides all the accelerations: the bucket's downward acceleration and the cylinder's rotational accleration (torque = tangential force x radius of cylinder)

mg - T= ma = F_{bucket}

Torque = I_{cyl}\alpha

alpha = a_{bucket}r

How do I solve this problem now
 

Answers and Replies

  • #2
Doc Al
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MAPgirl23 said:
How do I solve this problem now
By applying Newton's 2nd law to both the bucket and the cylinder, just like you are doing.

You've done it for the falling bucket: mg - T= ma. And you've almost done it for the cylinder: [itex]\tau = I \alpha[/itex]. But what is the torque?

What force exerts the torque on the cylinder? What's the rotational inertia of the cylinder?
 
  • #3
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knowing what I have, how do I solve for the tension in the rope?
 
  • #4
siddharth
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As Doc Al said,
In your equation [itex]\tau = I \alpha[/itex],
What is [itex]\tau[/itex]? (Remember [itex]\tau = RXF[/itex])Also, what is the value of "I" for a cylinder in this case?

Substitute those values and you will be able to find the value of acceleration and Tension from your equations
 
  • #5
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I for a cylinder is I = 1/2 x M x R^2

where R = 0.125 and M of the cylinder = 11.4 kg

but how do I find alpha = a_{bucket}r?
 
  • #6
siddharth
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You have 3 equations,
mg - T= ma -I
Torque = I_{cyl}\alpha -II
r(alpha) = a_{bucket} -III

and you have T,a,alpha as 3 unknowns.
What is the value of Torque? [Not I x alpha, thats the RHS. What is the LHS?]
 
Last edited:
  • #7
MAPgirl23 said:
A bucket of water of mass 14.9 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.250 m with mass 11.4 kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 10.2 m to the water. You can ignore the weight of the rope.

What is the tension in the rope while the bucket is falling?
With what speed does the bucket strike the water?
What is the time of fall?
While the bucket is falling, what is the force exerted on the cylinder by the axle?

** The gravitational force on the dropping bucket provides all the accelerations: the bucket's downward acceleration and the cylinder's rotational accleration (torque = tangential force x radius of cylinder)

mg - T= ma = F_{bucket}

Torque = I_{cyl}\alpha

alpha = a_{bucket}r

How do I solve this problem now
i think you have one of the equations wrong. It's a(bucket) = tangental acceleration of the windlass(alpha*r)

Draw a diagram. It helps a lot. Remember, the tangental force is the Force of tension on the windlass. And you have the equations for torque...
 
Last edited:
  • #8
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so to find the tension first of the rope I use mg - T = ma - I

where I_{cyl}= 0.5*M*R^2 = 0.5*11.4 kg*0.125^2 = 8.91 x 10^-2

ma - I = (11.4 kg)(9.8 m/s^2) - (8.91 x 10^-2) = 111.63

so, mg - T = 111.63 therefore T = mg - 111.63

T = (14.9 kg)(9.8 m/s^2) - 111.63 = 34.4 N
is that correct?
 
  • #9
MAPgirl23 said:
so to find the tension first of the rope I use mg - T = ma - I

where I_{cyl}= 0.5*M*R^2 = 0.5*11.4 kg*0.125^2 = 8.91 x 10^-2

ma - I = (11.4 kg)(9.8 m/s^2) - (8.91 x 10^-2) = 111.63

so, mg - T = 111.63 therefore T = mg - 111.63

T = (14.9 kg)(9.8 m/s^2) - 111.63 = 34.4 N
is that correct?
nope that's not correct. And how did the Inertia get into the equation of Mg - T = ma? Besides the acceleration of the bucket is not 9.8 because of the tension force from the rope.
 
  • #10
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then how do I find the acceleration to get the tension?
 
  • #11
Doc Al
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MAPgirl23 said:
then how do I find the acceleration to get the tension?
You have three equations and three unknowns:

[tex]\mbox(1) \ mg - T = ma[/tex]

[tex]\mbox(2) \ \tau = I \alpha \ \Longrightarrow \ rT = (1/2 M r^2) \alpha[/tex]

[tex]\mbox(3) \ a = r \alpha[/tex]

Solve them any way you want! Use equation 3 to eliminate alpha in equation 2; then use equation 2 and 1 together to solve for a and T.
 

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