A solid sphere of mass 2.5 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15° and rolls to the bottom. The upper end of the ramp is 1.2 m higher than the lower end. What is the linear speed of the sphere when it reaches the bottom of the ramp? (Note: I = 0.4MR2 for a solid sphere, and g = 9.8 m/s2.)

Question

BlackTom AI · Accepted Answer

We start by recognizing energy conservation for a solid sphere rolling without slipping down an incline. The gravitational potential energy mgh at the top converts into translational kinetic energy (1/2)mv^2 plus rotational kinetic energy (1/2)Iω^2 at the bottom, with v = ωR.

First, compute the rotational contribution using I = 0.4 m R^2 for a solid sphere:
(1/2)Iω^2 = (1/2)(0.4 m R^2)(v^2/R^2) = 0.2 m v^2.
Thus total kinetic energy K = (1/2)mv^2 + 0.2 m v^2 = 0.7 m v^2.

Set mgh equal to K: m g h = 0.7 m v^2. The mass m cancels, leaving g h = 0.7 v^2, hence v^2 = (g h) / 0.7.
Plug in g = 9.8 m/s^2 and h = 1.2 m: g h = 9.8 × 1.2 = 11.76.
So v^2 = 11.76 / 0.7 ≈ 16.8, and v ≈ √16.8 ≈ 4.10 m/s.

Now evaluate each option:
- 2.8 m/s: This is too small; it would correspond to a much smaller kinetic energy, ignoring the rotational contribution properly. The energy partition (0.7 factor) already accounts for rotation, so this undervalues v.
- 3.3 m/s: Still below the correct calculation; would imply a larger energy loss to rotation than allowed by the 0.7 factor given I = 0.4 mR^2. It does not satisfy mgh = 0.7 m v^2.
- 4.1 m/s: This matches the derived value v ≈ 4.10 m/s from the energy balance with rotational inertia included.
- none of these: Not correct because we found a consistent value that matches one of the options.
- 4.7 m/s: This would require a larger velocity than permitted by the energy distribution to translation and rotation, contradicting the 0.7 m v^2 kinetic term derived from I = 0.4 mR^2.

Summarizing, applying conservation of energy with the rotational inertia taken into account yields v = sqrt((g h)/0.7) ≈ 4.1 m/s, which aligns with option 4.1 m/s.

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