A solid sphere of mass 1.0 kg and radius 0.010 m starts from rest and rolls without slipping down a 1.0 m high inclined plane. What is the speed of the sphere when it reaches the bottom of the inclined plane?

Question

BlackTom AI · Accepted Answer

Use conservation of mechanical energy including rotational kinetic energy: the sphere’s loss of gravitational potential mgh becomes translational plus rotational kinetic energy. For a solid sphere the moment of inertia about its center is I = (2/5) m r^2 and rolling without slipping gives ω = v/r.

1) Relevant concepts
- Energy conservation: m g h = (1/2) m v^2 + (1/2) I ω^2.
- Moment of inertia for a solid sphere: I = (2/5) m r^2.
- Rolling without slipping: ω = v/r (so rotational KE can be written in v).

2) Step-by-step reasoning and algebra
m g h = (1/2) m v^2 + (1/2) I ω^2
Substitute I and ω:
m g h = (1/2) m v^2 + (1/2) * (2/5 m r^2) * (v^2 / r^2)

3) Intermediate calculation
The r^2 cancels in the rotational term:
m g h = (1/2) m v^2 + (1/5) m v^2
Combine coefficients:
(1/2 + 1/5) m v^2 = (5/10 + 2/10) m v^2 = (7/10) m v^2

So m g h = (7/10) m v^2
Cancel m and solve for v^2:
v^2 = (10/7) g h

Plug in g = 9.8 m/s^2 and h = 1.0 m:
v^2 = (10/7) * 9.8 = 98/7 = 14
v = sqrt(14) ≈ 3.7417 m/s ≈ 3.7 m/s

4) Why this answer is correct
The calculation accounts for both translational and rotational energy for a solid sphere; the factor 10/7 arises from the sphere’s moment of inertia (2/5 m r^2) and the rolling constraint. The radius cancels, so the small given radius does not change the speed.

5) Why the other options are incorrect
- 4.4 m/s is the speed for a frictionless block sliding (pure translation) from height 1.0 m: v = sqrt(2 g h) ≈ 4.43 m/s, which ignores rotational energy.
- 5.6 m/s and 6.3 m/s are significantly larger than allowed by energy conservation when rotation is included; they overestimate the translational kinetic energy and are not consistent with the solid-sphere moment of inertia.

6) Conclusion
The computed speed is ≈ 3.74 m/s, so the correct multiple-choice answer is 3.7 m/s, matching the provided answer.

A solid sphere of mass 1.0 kg and radius 0.010 m starts from rest and rolls without slipping down a 1.0 m high inclined plane. What is the speed of the sphere when it reaches the bottom of the inclined plane?Single choice

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