Calculating Tension and Velocity of Falling Bucket | Physics Problem Solution

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In summary: T = maT = (14.9 kg)(9.8 m/s^2) - 111.63 = 34.4 N\mbox(2) \ \tau = I \alpha\mbox(3) \ a = r \alpha\mbox(2) \ \tau = I \alpha\mbox(3) \ a = r \alpha\mbox(1) \ mg - T = maT = (14.9 kg)(9.8 m/s^2) - 111.63 = 34.4 N\
  • #1
MAPgirl23
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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
 
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  • #2
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
knowing what I have, how do I solve for the tension in the rope?
 
  • #4
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
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
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, that's 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
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
then how do I find the acceleration to get the tension?
 
  • #11
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.
 

What is the concept of "Falling bucket"?

The concept of "Falling bucket" refers to the study of how a bucket filled with water falls from different heights and how the water inside the bucket behaves during the fall.

What factors affect the behavior of a falling bucket?

The factors that affect the behavior of a falling bucket include the height from which the bucket is dropped, the amount of water in the bucket, the shape and size of the bucket, and the force of gravity.

What happens to the water inside the bucket during the fall?

As the bucket falls, the water inside experiences a downward force due to gravity. This causes the water to accelerate and push against the bottom of the bucket, creating a pressure difference. As a result, the water rises up the sides of the bucket and can even splash out if the force is strong enough.

How does air resistance affect a falling bucket?

Air resistance, also known as drag, can affect the speed and trajectory of a falling bucket. It opposes the motion of the bucket, slowing it down and causing it to fall at a slower rate. This can also impact the behavior of the water inside the bucket, as it may experience turbulence due to air resistance.

What are the real-world applications of studying "Falling bucket"?

The study of "Falling bucket" has various real-world applications, such as in understanding fluid dynamics and air resistance. It can also be applied in engineering and design, for example, in the development of better parachutes or water tanks. Additionally, it can help in predicting and preventing potential hazards, such as water splashing out of a bucket during transportation.

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