Question

# A bicycle wheel makes revolutions per minute to maintain a speed of 8.91 km per hour, find the diameter of the wheel.

Hint: The problem is related to tangential and angular speed. The relation between tangential and angular speed is, $v = r\omega$ .

Complete step by step solution: Let’s assume the number of revolutions per minute be N.
The bicycle speed is given as
$v = 8.91$ km per hour.
This is the tangential speed of the bicycle wheel. It is the speed which is measured at any point tangent to the wheel. Its unit is km per hour.
To convert speed in km per hour to m per minute, the factor to be multiplied is, $\dfrac{{1000}}{{60}} = \dfrac{{50}}{3}$ .
Bicycle speed (in meter per minute),
$v = 8.91 \times \dfrac{{50}}{3} \\ v = 148.5 \\$
The relation between the wheels tangential speed, angular speed and the wheel radius is given by,
Tangential speed = Radius of the wheel x Angular Speed
$v = r\omega \cdots (1)$
The angular speed is the rate at which the object changes its angle and is measured in radians per second. In terms of revolutions per minute is given by,
$\omega = \dfrac{{2\pi N}}{{60}} \cdots (2)$
Where, The number of revolutions made by the wheel in one minute.
Substituting the value of in equation (1),
$v = r \times \dfrac{{2\pi N}}{{60}} \cdots (3)$
The radius of the wheel is calculated by rearranging the terms in equation (3),
$r = \dfrac{{60v}}{{2\pi N}} \cdots (4)$
The diameter of the wheel is twice its radius, put $r = \dfrac{d}{2}$ in equation (4),
$\dfrac{d}{2} = \dfrac{{60v}}{{2\pi N}} \\ d = \dfrac{{60v}}{{\pi N}} \\ d = \dfrac{{60v}}{{\pi N}} \cdots (5) \\$
Put the value of bicycle speed as $v = 148.5$ meter per minute in equation (5),
$d = \dfrac{{60 \times 148.5}}{{\pi N}} \\ d = \dfrac{{2836.14}}{N} \\$
Thus, the diameter of the wheel is depending on the value of the number of revolutions made by the bicycle wheel per minute.

Note: The relation between the angular speed and tangential speed comes from the concept of circular motion. As the value of tangential speed increases more number of revolutions will take place, as result of which angular velocity increases.
The following relations between the tangential speed , angular speed and radius of the wheel should be remembered
$v = r\omega$
Also, $\omega = \dfrac{{2\pi N}}{{60}}$ , where N= number of revolutions per minute.