How Does Mass and Height Affect the Angular Speed of a Rotating Bicycle Wheel?

In summary, the angular speed of the wheel can be found by using the equation ω = √[(2mgh - mv^2)/I], where m is the mass of the block, g is the acceleration due to gravity, h is the distance the block falls, r is the radius of the wheel, and I is the moment of inertia about its rotation axis. Once solved, you can replace either v or ω with the correct relationship, such as ω = v^2/r, to find the final solution.
  • #1
PascalPanther
23
0
"Consider the bicycle wheel is not turning initially. A block of mass m is attached (with a massless string) to a wheel. The block is allowed to fall a distance of h. Assume that the wheel has a moment of inertia I about its rotation axis.

Find the angular speed of the wheel after the block has fallen a distance of h in terms of m,g,h, r(of the wheel) and I"

This is what I did:

U + K + W(other) = U + K
mgh + 0 + 0 = 0 + 1/2mv^2 + 1/2 I*omega^2

omega = Sqrt[(2mgh - mv^2)/I]

This is wrong... what am I doing wrong? I know I can use omega = v^2/r, but not sure where that would go...
 
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  • #2
PascalPanther said:
"Consider the bicycle wheel is not turning initially. A block of mass m is attached (with a massless string) to a wheel. The block is allowed to fall a distance of h. Assume that the wheel has a moment of inertia I about its rotation axis.

Find the angular speed of the wheel after the block has fallen a distance of h in terms of m,g,h, r(of the wheel) and I"

This is what I did:

U + K + W(other) = U + K
mgh + 0 + 0 = 0 + 1/2mv^2 + 1/2 I*omega^2

omega = Sqrt[(2mgh - mv^2)/I]

This is wrong... what am I doing wrong? I know I can use omega = v^2/r, but not sure where that would go...
ω = v^2/r is not correct, but it is sort of close. When you get it right, you can replace either v or ω using the correct relationship. Since you want to solve for ω, replace the v and solve away.
 
  • #3


Your approach is on the right track, but there are a few errors in your calculation. Let's break down the steps and correct them:

1. Identify the initial and final states of the system: In this case, the initial state is when the wheel is not turning and the block is at a height h above the ground. The final state is when the block has fallen to the ground and the wheel is rotating.

2. Write the conservation of energy equation: As you correctly did, the total energy of the system (U + K + W) in the initial state must equal the total energy in the final state.

3. Calculate the initial and final potential energies: In the initial state, the potential energy is mgh, as you correctly identified. In the final state, the potential energy is 0, since the block is now at the ground.

4. Calculate the initial and final kinetic energies: In the initial state, the kinetic energy is 0 since the wheel is not rotating. In the final state, the kinetic energy is 1/2mv^2 for the block and 1/2Iω^2 for the wheel, as you correctly identified.

5. Set the initial and final energies equal to each other: This is where your calculation went wrong. You should set mgh = 1/2mv^2 + 1/2Iω^2, since the total energy must be conserved.

6. Use the relationship between linear and rotational speed: As you correctly mentioned, the linear speed of the block is related to the angular speed of the wheel by v = ωr, where r is the radius of the wheel. Substituting this into the equation from step 5, we get mgh = 1/2m(ωr)^2 + 1/2Iω^2.

7. Solve for ω: Rearranging the equation from step 6, we get ω = √[2mgh/(mr^2 + I)]. This is the correct equation for the angular speed of the wheel after the block has fallen a distance of h.

In summary, your approach was correct, but there were a few errors in your calculation. By carefully identifying the initial and final states, writing the conservation of energy equation, and correctly substituting the relationship between linear and rotational speed, we arrive at the correct equation for the angular speed of the wheel.
 

Related to How Does Mass and Height Affect the Angular Speed of a Rotating Bicycle Wheel?

1. What is rotational kinetic energy?

Rotational kinetic energy is the energy possessed by a rotating object due to its motion. It is dependent on the object's mass, rotational speed, and the distribution of its mass around the axis of rotation.

2. How is rotational kinetic energy different from linear kinetic energy?

Rotational kinetic energy is different from linear kinetic energy in that it involves the rotation of an object around an axis, while linear kinetic energy is associated with the motion of an object in a straight line.

3. What is the formula for calculating rotational kinetic energy?

The formula for rotational kinetic energy is KE= 1/2 * I * ω^2, where KE is the kinetic energy, I is the moment of inertia of the object, and ω is the angular velocity.

4. How does rotational kinetic energy affect an object's stability?

Rotational kinetic energy can affect an object's stability by influencing its ability to maintain its orientation and resist changes in its rotational motion. Objects with higher rotational kinetic energy are generally less stable.

5. What are some real-life examples of rotational kinetic energy?

Some real-life examples of rotational kinetic energy include a spinning top, a basketball rotating on a player's finger, and a wind turbine rotating to generate electricity. Other examples include the spinning motion of a planet around its axis and the rotation of a propeller on an airplane.

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