Question

A large 3kg object hangs from a rope wound on a 40 kg wheel. The wheel has an actual radius of 0.75m and a radius of gyration of 0.60m. Find (a) the angular acceleration and (b) the distance through which the weight will fall in the first 10s.

Answer 1

Answer on Question #55265, Physics / Mechanics | Kinematics | Dynamics

A large $3\mathrm{kg}$ object hangs from a rope wound on a $40\mathrm{kg}$ wheel. The wheel has an actual radius of $0.75\mathrm{m}$ and a radius of gyration of $0.60\mathrm{m}$. Find (a) the angular acceleration and (b) the distance through which the weight will fall in the first 10s.

Solution:

(a) The net torque is

$$\tau = F r _ {1}$$

where $r_1 = 0.75 \, \text{m}$ and $F = m_1 g = (3 \, kg) * (9.8 \, \text{m/s}^2) = 29.4 \, \text{N}$

Newton’s second law for rotation:

$$\tau = I \alpha$$

where $I$ is moment of inertia and $\alpha$ is angular acceleration.

The radius of gyration about a given axis $r_g$ can be computed in terms of the mass moment of inertia $I$ around that axis, and the total mass $m$;

$$r _ {g} = \sqrt {\frac {I}{m _ {2}}}$$

Thus,

$$I = m _ {2} r _ {g} ^ {2}$$

Hence,

$$\alpha = \frac {\tau}{I} = \frac {m _ {1} g r _ {1}}{m _ {2} r _ {g} ^ {2}} = \frac {3.0 * 9.8 * 0.75}{40 * 0.60^2} = 1.53 \, \mathrm{rad/s^2}$$

(b) The relationship between acceleration of the rope, $a$, and the angular acceleration, $\alpha$

$$a = r _ {1} \alpha$$

The distance through which the weight will fall is

$$h = \frac {a t ^ {2}}{2} = \frac {r _ {1} \alpha t ^ {2}}{2} = \frac {0.75 * 1.53 * 10^2}{2} \approx 57.4 \, \text{m}$$

Answer: (a) $1.53 \, \mathrm{rad/s^2}$; (b) $57.4 \, \text{m}$

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