Consider a wheel to be a ring of radius 30 cm and mass 1.5 kg. A force F is applied tangentially to a sprocket of radius 4.0 cm, causing the wheel to accelerate to a speed of 15 m/s over a distance of 50 m. Assuming the wheel rolls without slipping, calculate F.(Hint: because the wheel rotates as it moves, the work done by F is not just F x 50 m.)单项选择题
A
6.0 N
B
3.0 N
C
4.5 N
D
6.8 N

Question

Consider a wheel to be a ring of radius 30 cm and mass 1.5 kg. A force F is applied tangentially to a sprocket of radius 4.0 cm, causing the wheel to accelerate to a speed of 15 m/s over a distance of 50 m. Assuming the wheel rolls without slipping, calculate F.(Hint: because the wheel rotates as it moves, the work done by F is not just F x 50 m.)单项选择题

A

6.0 N

B

3.0 N

C

4.5 N

D

6.8 N

BlackTom AI · Accepted Answer

Energy balance and rolling constraints determine F. Because the wheel rolls without slipping, the work done by the applied tangential force must supply both the translational kinetic energy of the wheel’s center of mass and the rotational kinetic energy about the centre.

1) Relevant concepts
- Work-energy theorem: work done by external force = change in total kinetic energy (translational + rotational).
- For a thin ring: I = mR^2. Translational KE = (1/2) m v^2. Rotational KE = (1/2) I ω^2 with ω = v/R for rolling without slipping.
- If the force is applied tangentially to a sprocket rigidly attached to the wheel, the point at which the force acts undergoes the same angular displacement θ as the wheel. The linear distance that point moves while the wheel’s centre moves by s is L = r θ = r (s/R).

2) Step-by-step reasoning
- Compute translational KE: (1/2) m v^2 = 0.5 × 1.5 × (15)^2 = 0.5 × 1.5 × 225 = 168.75 J.
- Compute rotational KE using I = mR^2 and ω = v/R:
  I = m R^2 = 1.5 × (0.30)^2 = 1.5 × 0.09 = 0.135 kg·m^2.
  ω = v/R = 15 / 0.30 = 50 rad/s.
  Rotational KE = (1/2) I ω^2 = 0.5 × 0.135 × 50^2 = 0.5 × 0.135 × 2500 = 168.75 J.
- Total kinetic energy gained = translational + rotational = 168.75 + 168.75 = 337.5 J.

3) Work done by F
- The force F acts tangentially on the small sprocket; the point of application moves a distance L = (r/R) s = (0.04/0.30) × 50 = 6.6667 m while the wheel’s centre moves s = 50 m.
- The work done by F is W = F × L. Set this equal to the increase in total kinetic energy:
  F × L = 337.5 J → F = 337.5 / 6.6667 = 50.625 N.

4) Why the (intended) correct answer is the given choice
- The correct application of energy and rolling kinematics shows the work by F must provide both translational and rotational kinetic energy, and because the force acts on the smaller sprocket the force moves through a smaller distance L = (r/R)s. Using those relations gives the force required to achieve v = 15 m/s over s = 50 m.
- Note: if instead the shorter form F s = total KE were (incorrectly) used one would get F ≈ 6.75 N (~6.8 N). The full, physically consistent calculation with the sprocket motion gives F ≈ 50.6 N.

5) Why the other options are incorrect
- 3.0 N and 4.5 N are far too small — they correspond to neglecting rotational energy or misallocating energy between rotation and translation.
- 6.8 N would match the mistaken calculation that takes W = F × 50 m = total KE (ignoring that the force acts on the small sprocket and moves a shorter distance).
- The physically consistent calculation (above) gives ≈50.6 N, so the listed options do not match that rigorous result; among the multiple-choice answers 6.8 N would correspond to the common mistaken simplification.

Conclusion: Using energy conservation with the sprocket-to-wheel geometry yields the force required as about 50.6 N. If one follows the simplified (incorrect) assumption that the force moves through the full 50 m one instead finds ≈6.8 N (the closest listed option). The provided answer listed is 6.0 N.