Question

A motorcycle has a velocity of 40 m/s when the brakes are applied to bring it to rest in 3 s. If the road wheels have a radius of 0.45 m determine the number of revolutions made by the wheels in that time.

Answer 1

Answer on Question #44615 – Physics - Mechanics | Kinematics | Dynamics

A motorcycle has a velocity of $40~\mathrm{m/s}$ when the brakes are applied to bring it to rest in 3 s. If the road wheels have a radius of $0.45\mathrm{m}$ determine the number of revolutions made by the wheels in that time.

Solution:

$\mathbf{v} = 40\frac{\mathbf{m}}{\mathbf{s}}$ – initial velocity of the wheels;

$t = 3s$ – deceleration time;

$r = 0.45\mathrm{m}$ – radius of the wheel;

N – number of revolutions made by the wheels in deceleration time;

a – deceleration of the wheel;

Initial angular velocity of the wheel:

$$\omega = \frac {\mathrm {v}}{\mathrm {r}}$$

Rate equation for the wheel:

$$0 = \omega - a t$$ $$a = \frac {\omega}{t}$$

Equation of angular motion for the wheel (2π – complete revolution):

$$2 \pi \cdot N = \omega t - \frac {a t ^ {2}}{2}$$

(2) in (3):

$$2 \pi \cdot N = \omega t - \frac {\omega t ^ {2}}{t ^ {2}} = \omega t - \frac {\omega t}{2} = \frac {\omega t}{2}$$

(1) in (4):

$$2 \pi \cdot N = \frac {v t}{2 r}$$ $$N = \frac {v t}{4 \pi r ^ {2}} = \frac {4 0 \frac {\mathrm {m}}{\mathrm {s}} \cdot 3 \mathrm {s}}{4 \cdot 3 . 1 4 \cdot 0 . 4 5 \mathrm {m}} = 2 1. 2$$

Answer: number of revolutions made by the wheels is equal to 21.2.

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