Looking at the experimental setup, write down an equation for the total mechanical energy E of the system just as the load reaches the ground. This should be in terms of the gravitational acceleration g and the following variables; the velocity v of the load, the angular velocity w of the wheel, the moment of inertia I (capital i) of the wheel, the initial height H of the wheel’s centre of mass, the mass m of the load, and the mass M of the wheel.

For your answer, please follow the following:
Please write the answer without any blanks.
For powers use ^ ie for v2 use v^2.
Please use the common order for the letters, as used in the lectures/textbook ie for kinetic energy E=1/2mv^2 (not E=1/2v^2m)
Write your answer in the following form:
E=1/2(ab+cd^2)
(your answer might not be in this form, this is just to demonstrate how multiplication, division, addition, power,... should be written. Note: only factorise numbers not variables.)

Question

Looking at the experimental setup, write down an equation for the total mechanical energy E of the system just as the load reaches the ground. This should be in terms of the gravitational acceleration g and the following variables; the velocity v of the load, the angular velocity w of the wheel, the moment of inertia I (capital i) of the wheel, the initial height H of the wheel’s centre of mass, the mass m of the load, and the mass M of the wheel.

For your answer, please follow the following:
Please write the answer without any blanks.
For powers use ^ ie for v2 use v^2.
Please use the common order for the letters, as used in the lectures/textbook ie for kinetic energy E=1/2mv^2 (not E=1/2v^2m)
Write your answer in the following form:
E=1/2(ab+cd^2)
(your answer might not be in this form, this is just to demonstrate how multiplication, division, addition, power,... should be written. Note: only factorise numbers not variables.)

BlackTom AI · Accepted Answer

Mechanical energy is the sum of translational kinetic energy, rotational kinetic energy, and gravitational potential energy, so write E = K_trans + K_rot + U_grav.

1) Relevant knowledge points and concepts
- Translational kinetic energy of the load: K_trans = 1/2 m v^2, where m is the load mass and v its speed as it reaches the ground.
- Rotational kinetic energy of the wheel: K_rot = 1/2 I w^2, where I is the wheel’s moment of inertia and w its angular speed.
- Gravitational potential energy of the wheel’s centre of mass: U = MgH, taking the ground as the zero of potential and H as the initial height of the wheel’s centre of mass. The load’s potential energy at the instant it reaches the ground is zero.
- Any translational kinetic energy of the wheel’s centre of mass is assumed absent (the wheel rotates about its axis without net translation), so no 1/2 M v^2 term for the wheel is included.

2) Step-by-step reasoning
- Total mechanical energy at the instant the load hits the ground equals the instantaneous kinetic energies plus the remaining potential energy of the wheel:
  E = K_trans(load) + K_rot(wheel) + U_grav(wheel).
- Substitute the standard expressions for each term:
  K_trans(load) = 1/2 m v^2.
  K_rot(wheel) = 1/2 I w^2.
  U_grav(wheel) = MgH (since the wheel’s CM is at height H above the ground reference).

3) Intermediate steps / algebra
- Combine the three terms:
  E = 1/2 m v^2 + 1/2 I w^2 + MgH.

4) Why this is correct
- Each term corresponds to an independent form of mechanical energy present at that instant: the load’s translational KE, the wheel’s rotational KE, and the wheel’s gravitational PE. The load’s gravitational PE is zero at ground, and no additional wheel translational KE is given.

5) Common incorrect alternatives explained briefly
- Adding m g H would be wrong because the load is at ground (its potential is zero at this instant).
- Adding 1/2 M v^2 would be wrong unless the wheel’s centre of mass were translating; the problem gives only rotation (I, w) for the wheel.

6) Conclusion
This leads to E=1/2mv^2+1/2Iw^2+MgH, which matches the provided correct answer.

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