Time Dependent Torque/Merry Go Round

In summary, Given an uniform merry go round with solid disc radius R and mass M1, set spinning with initial angular velocity omega naut and stops after time t1, the magnitude of the constant frictional torque is 53.38N. With a time dependent force tangent to the merry go round given by equation F(t) = Fo*e^(-bt), exerted for time t2 and then jumping onto the edge of the merry go round, the angular velocity after stop pushing is 5.381 rad/s. After jumping onto the merry go round, the speed is 12.38m/s. To find alpha, use Newton's 2nd law for rotational motion: \tau = I \alpha. To find the
  • #1
twiztidmxcn
43
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So, you have an uniform merry go round with solid disc radius R and mass M1. It is set spinning with initial angular velocity of omega naut. If it stops after time t1, find magnitude of constant frictional torque that slows it down.

M1 = 243 kg, R = 2.3m, omega nau = 2.5 rad/s, t1 = 30.1 s

So i found frictional torque to be 53.38N. The next part states that you exert a time dependent force tangent to the merry go round as run along side it. This is given by equation F(t) = Fo*e^(-bt) where Fo is F naut and is a constant and b is a constant. You exert the force for time t2 and then jump onto edge of the merry go round and because you move with same speed as a point on the edge of the merry go round, when you jump onto the edge you don't affect its angular velocity. If frictional torque is same throughout entire process, find angular velocity after stop pushing.

Fo = 402N, b = 0.18 s^-1, t2 = 7.8s, your mass is 87kg. So i found omega = 5.381 rad/s.

What was my speed after jump on the merry go round? v = 12.38m/s

Now, here's where the problem is. How many revolutions do you make on the merry go round before it stops? Your mass is 87kg. The frictional torque is the same throughout entire process and when you jump onto the edge of the merry go round, you change its moment of inertia.

I know omega, frictional torque and moment of inertia and that I can find alpha somehow. But how do I find alpha from knowing that information, and how can I find the time until it stops and how many revolutions there are?

I am completely lost at this point, I have a ton of numbers and equations but cannot find something that gives me the right answer. Any help would be appreciated.
 
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  • #2
twiztidmxcn said:
I know omega, frictional torque and moment of inertia and that I can find alpha somehow. But how do I find alpha from knowing that information,
By applying Newton's 2nd law for rotational motion: [itex]\tau = I \alpha[/itex].
and how can I find the time until it stops and how many revolutions there are?
Use the kinematic equations for uniformly accelerated motion. You know the initial angular speed and the angular acceleration: Use the rotational analog of [itex]v_f^2 = v_0^2 + 2as[/itex].
 
  • #3


It seems that you have all the necessary information to solve this problem, but you are just unsure of how to use it to find the number of revolutions and time until the merry go round stops. Here are some steps that may help you:

1. Start by finding the moment of inertia of the merry go round with the person on the edge. This can be done by using the parallel axis theorem, which states that the moment of inertia of a body is equal to the moment of inertia about its center of mass plus the product of its mass and the square of the distance between the two axes. In this case, the center of mass will be at the center of the merry go round, and the person will be at a distance equal to the radius of the merry go round.

2. Once you have the moment of inertia, you can use the equation τ = Iα to find the angular acceleration (α) of the merry go round. Remember that the frictional torque (τ) is equal to the product of the frictional force and the radius of the merry go round.

3. With the angular acceleration, you can use the equations of motion for rotational motion to find the angular velocity (ω) at any given time. In this case, you will need to use the initial angular velocity (ω0), the angular acceleration (α), and the time (t) to find ω at the end of the time period t1 (when the merry go round stops spinning).

4. Now that you have the angular velocity at the end of t1, you can use the equation ω = Δθ/Δt to find the number of revolutions (Δθ) that the merry go round made during the time t1.

5. To find the time until the merry go round stops, you can use the equation ω = ω0 + αt to find the time (t) at which ω will be equal to 0. This will give you the total time until the merry go round stops spinning.

I hope this helps guide you in the right direction. Remember to carefully consider all the given information and use the appropriate equations to solve the problem. Good luck!
 

1. What is time dependent torque?

Time dependent torque refers to a torque force that changes over time. This means that the amount or direction of the torque is not constant, but instead varies as time passes.

2. How does time dependent torque affect a merry go round?

Time dependent torque can affect a merry go round by causing it to rotate at a varying speed. This is because the force of the torque is not constant, so the speed of rotation will also change.

3. What causes time dependent torque?

Time dependent torque can be caused by various factors, such as changes in the external forces acting on an object or changes in the object's shape or composition over time.

4. How is time dependent torque measured?

Time dependent torque can be measured using a torque sensor, which can detect changes in the amount and direction of torque acting on an object. This sensor can then provide data on the torque over a specific period of time.

5. What are some real-world applications of time dependent torque?

Time dependent torque has many practical applications, such as in the operation of machinery and vehicles, the motion of celestial bodies, and the analysis of fluid dynamics. It is also important in understanding the behavior of rotating systems and the effects of changing forces on their movement.

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