Uniform circular motion is a motion of a body of mass 𝑚 in a circular path of constant radius 𝑟 at constant speed. The velocity 𝑣 changes and is always tangent to the circle but its magnitude is constant. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant as shown in figure below.
In this post we will take a deep look into circular motion by solving many problems. A body in uniform circular motion undergoes at all times a centripetal acceleration which is directed towards the center of motion.
Question 1
A 5 kg object moves around a circular track of a radius of 18 cm with a constant speed of 6 m/s.
Find (a) The magnitude and direction of the acceleration of the object. (b) The net force acting upon the object causing this acceleration.
Show Solution
mass, m = 5kg
radius, r = 18cm = 0.18 m
velocity, v = 6 m/s
(a) The centripetal acceleration, \(\:a_r \:\) is given by
$$a_r=\frac{v^2}{r} = \frac{6^2}{0.18}=200 \:ms^{-2}$$
The direction of the acceleration is towards the center
(b)
The net force acting, F is given by
$$F=ma_r=5\times 200 = 1000 \:N$$
Question 2
A 20 kg child sitting in a cart to which a 2 m rope is attached. The rope is tied to a motor that rotates the cart. At the instant that the tension in the rope is 100 N, how many revolutions per minute does the cart make?
Show Solution
mass, m = 20 kg
radius, r = 2 m
centripetal force, F = tension in the rope = 100N
The centripetal force \(F\) is given by
$$F=\frac{mv^2}{r} = m\omega^2 r$$
$$\omega^2=\frac{F}{mr} = \frac{100}{20 \times 2} =2.5$$
$$\omega=\sqrt{2.5} = 1.58 \: rads^{-1}$$
$$1 \: radian=\frac{1}{2\pi} \: \text{revolutions}=0.159 \:rev$$
$$1 \: second=\frac{1}{60} \: \text{minutes}=0.017 \: min$$
$$\text{Therefore,}\: \omega=\frac{1.58 \times 0.159}{0.017} \: rev/min=14.79 \: rev/min$$
Question 3
Calculate the angular speed of the second hand, minute hand and hour hand of a clock.
Show Solution
In every 1 second, the second hand of the clock rotates \(360^0\).
Thus, the angular velocity \(\omega\) for a second hand will be given by
$$\omega=\frac{\theta}{t}=\frac{360^0}{1}=\frac{2\pi}{1}=2\pi=6.28 \: rad/s$$
In every 1 minute = 60 seconds, the minute hand of the clock rotates \(360^0\).
Thus, the angular velocity \(\omega\) for a minute hand will be given by
$$\omega=\frac{\theta}{t}=\frac{360^0}{60}=\frac{2\pi}{60}=0.105 \: rad/s$$
In every 1 hour = 3600 seconds, the hour hand of the clock rotates \(360^0\).
Thus, the angular velocity \(\omega\) for a hour hand will be given by
$$\omega=\frac{\theta}{t}=\frac{360^0}{3600}=\frac{2\pi}{3600}=0.00174 \: rad/s$$
Question 4
A particle is traveling in a circle of radius of 2.5 m and with an angular velocity of 10 rad/s.
The particle begins to slow down with an angular acceleration of -1 rad/s2. After 5 seconds, what is the centripetal acceleration
Show Solution
radius, r = 2.5 m
Initial angular velocity,\(\: \omega_0\) = 10 rad/s
Angular acceleration,\(\: \alpha\) = -1 \(rad/s^2\)
After time t = 5 seconds, the final angular velocity,\(\: \omega\) will be given by
$$\omega=\omega_0 + \alpha t=10 \: – 1 \times 5 = 5 \: rad/s$$
The centripetal acceleration \( a_r\) will be given by
$$a_r = \omega^2 r=5 \times 2.5 = 12.5 \: m/s^2$$
Question 5
To what angle must a racing track of radius of curvature 600 m be banked so as to be suitable for a maximum speed of 180 km/h?
Show Solution
radius, r = 600 m
speed, v = 180 km/h = 50 m/s
accelaration due to gravity, g = 9.82 \(\: m/s^2\)
The banked angle \(\: \theta \:\) is given by
$$\theta =tan^{-1}\left( \frac{v^2}{rg} \right) =tan^{-1}\left( \frac{50^2}{600 \times 9.82} \right) $$
$$=30^0$$
Question 6
A vehicle enters a circular bend of radius 200 m at 72 km/h. The road surface at the bend is banked at \(\: 10^0 \: \). Is it safe? At what angle should the road surface be ideally banked for safe driving at this speed?
Show Solution
radius, r = 200 m
accelaration due to gravity, g = 9.82 \(\: m/s^2\)
The banked angle, \(\: \theta = 10^0 \)
The maximum speed limit without skidding, v is given
$$v = \sqrt{rgtan\theta} = \sqrt{200 \times 9.82 \times tan10^0}$$
$$=18.61 \: m/s = \frac{18.61 \times 3600}{1000} \: km/h$$
$$67 \: km/h$$
Since the speed limit without skidding, 67 km/h which is less than the speed of 72 km/h with which the vehicle enters the circular bend, therefore it is not safe to drive with such speed of 72 km/h.
Let us now find the banked angle, \(\: \alpha \: \) for a speed of 72 km/h = 20 m/s = \(\: u \:\) as follows:
$$\alpha =tan^{-1}\left( \frac{u^2}{rg} \right) =tan^{-1}\left( \frac{20^2}{600 \times 9.82} \right) $$
$$=11.5^0$$
Therefore, the banked angle for a speed of 72 km/h is \(\: 11.5^0\)
Question 7
A crankshaft of radius 8 cm rotates at 2400 rpm (revolutions per minute). What is the speed of a point at the surface of it?
Show Solution
radius, r = 8 cm = 0.08 m
angular velocity, \(\: \omega = 2400 \: rev/min = \frac{2400 \times 2\pi}{60}\: rad/s = 251.4 \: rad/s\)
The linear speed \(v\) of a point at the surface is given by
$$v=\omega r = 251.4 \times 0.08 = 20.1 m/s$$
Question 8
A radially inward constant force of 300 N is exerted on a 2 kg ball as it revolves around a circle of radius 85 cm. What is the speed of the ball?
Show Solution
centripetal force, F = 300 N
mass, m = 2 kg
radius, r = 85 cm = 0.85 m
speed, \(\: v \: \) of the ball = ?
By defination,
$$F = \frac{mv^2}{r}$$
$$v = \sqrt{\frac{Fr}{m}}=\sqrt{\frac{300 \times 0.85}{2}} = 11.3 \: m/s$$
Question 9
A circular road course track has a radius of 500 m and is banked to \(\: 10^0 \:\). If the coefficient of friction between the road and the tyre is 0.25. Compute the maximum speed to avoid slipping.
Show Solution
radius, r = 500 m
accelaration due to gravity, g = 9.82 \(\: m/s^2\)
banked angle, \(\: \theta = 10^0\)
coeficient of friction, \(\:\mu = 0.25\)
The maximum speed limit without skidding, \(\: v \:\) is given
$$v = \sqrt{\frac{rg(tan\theta + \mu)}{1-\mu tan \theta}}$$
$$ = \sqrt{\frac{500 \times 9.82 \times (tan10^0 + 0.25)}{1-0.25 \times tan10^0}}$$
$$\sqrt{\frac{2093.3}{0.956}}=46.8 \: m/s$$
Question 10
What is the angular displacement of the minute hand of a clock in 25 minutes?
Show Solution
In every 1 minute, the minute hand of a clock rotates by \(\:360^0 = 2\pi \:\text{radians}\)
Thus, the angular displacement in 25 minutes = \(\: 25 \times 2\pi \: rad = 50\pi \:rad =157.1 \: rad\)
Question 11
What is the acceleration of a stone attached to the end of a 2 m long rope revolving at 45 revolutions per minute (rpm)?
Show Solution
radius, r = 2 m
angular velocity, \(\:\omega = 45 \: rev/min = \frac{45\times 2\pi}{60} \: rad/s = 4.71 \: rad/s\)
The centripetal acceleration, \(\: a_r \:\) is given by
$$a_r = \frac{v^2}{r} =\frac{4.71^2}{2} = 11.11 \: m/s^2$$
