Questions on circular motion

Question 1

Explain why an object moving at constant speed in a circle is nevertheless accelerating.

Accepted Answer

Velocity is a vector, so although the speed is constant the direction of motion is continually changing (1). A changing velocity means the object is accelerating, towards the centre of the circle (1).

Question 2

An object rotates at a frequency of 5.0 Hz. Calculate its angular speed.

Accepted Answer

ω = 2πf = 2π × 5.0 (1) ω = 31.4 rad s⁻¹ (1)

Question 3

An object moves in a circle of radius 2.0 m at a speed of 10 m s⁻¹. Calculate its angular speed.

Accepted Answer

ω = v/r = 10/2.0 (1) ω = 5.0 rad s⁻¹ (1)

Question 4

A car travels 24 m along a bend that is an arc of radius 60 m. Calculate the angle, in radians, through which the car turns.

Accepted Answer

θ = s/r = 24/60 = 0.40 rad (1).

Question 5

A drill bit rotates at 3000 revolutions per minute. Calculate its angular speed.

Accepted Answer

f = 3000/60 = 50 revolutions per second (1) ω = 2πf = 2π × 50 = 314 rad s⁻¹ (1)

Question 6

State what provides the centripetal force on (a) the Moon in its orbit round the Earth and (b) a car travelling round a flat roundabout.

Accepted Answer

(a) The gravitational attraction of the Earth on the Moon (1). (b) Friction between the tyres and the road surface (1).

Question 7

A 0.50 kg object moves in a circle of radius 4.0 m at 8.0 m s⁻¹. Calculate (a) its centripetal acceleration and (b) the centripetal force on it.

Accepted Answer

(a) a = v²/r = 8.0²/4.0 (1) a = 16 m s⁻² (1) (b) F = ma = 0.50 × 16 = 8.0 N (1)

Question 8

A car of mass 1000 kg goes round a bend of radius 50 m at 15 m s⁻¹. Calculate the centripetal force required.

Accepted Answer

F = mv²/r = (1000 × 15²)/50 (1) F = 4500 N (1)

Question 9

A point on the rim of a spinning wheel moves in a circle of radius 0.30 m at a frequency of 2.0 Hz. Calculate (a) its angular speed, (b) its linear speed and (c) its centripetal acceleration.

Accepted Answer

(a) ω = 2πf (1) ω = 12.6 rad s⁻¹ (1) (b) v = ωr = 3.77 m s⁻¹ (1) (c) a = ω²r = 47.4 m s⁻² (1)

Question 10

A 0.20 kg ball is whirled in a horizontal circle of radius 0.80 m at 4.0 m s⁻¹. Calculate the tension in the string providing the centripetal force.

Accepted Answer

T = mv²/r = (0.20 × 4.0²)/0.80 (1) T = 4.0 N (1)

Question 11

A student swings a bucket of water in a vertical circle of radius 0.90 m (g = 9.81 m s⁻²). Calculate the minimum speed at the top of the circle for the water to stay in the bucket.

Accepted Answer

At the minimum speed, the weight of the water alone provides the centripetal force at the top (1) mg = mv²/r, so v = √(gr) = √(9.81 × 0.90) (1) v = 3.0 m s⁻¹ (1)

Question 12

A coin of mass 12 g rests 0.15 m from the centre of a turntable rotating at 45 revolutions per minute. The maximum frictional force the surface can exert on the coin is 0.045 N. Deduce whether the coin stays in place or slides off.

Accepted Answer

ω = 2π × 45/60 = 4.71 rad s⁻¹ (1)
Force needed = mω²r = 0.012 × 4.71² × 0.15 (1)
Force needed = 0.040 N (1)
0.040 N is less than the 0.045 N available from friction, so the coin stays in place (1)

Question 13

A cyclist rides at constant speed round a circular track of radius 25 m, completing one lap in 18 s. Calculate the cyclist's speed and centripetal acceleration.

Accepted Answer

v = 2πr/T = 2π × 25/18 (1) v = 8.7 m s⁻¹ (1) a = v²/r = 8.7²/25 = 3.0 m s⁻² (1)

Question 14

A 0.10 kg object moves in a horizontal circle of radius 0.50 m with a period of 0.40 s. Calculate (a) its angular speed, (b) its centripetal acceleration and (c) the centripetal force on it.

Accepted Answer

(a) ω = 2π/T = 2π/0.40 (1) ω = 15.7 rad s⁻¹ (1) (b) a = ω²r = 123 m s⁻² (1) (c) F = ma = 0.10 × 123 = 12.3 N (1)

Question 15

A 0.25 kg mass is whirled in a horizontal circle of radius 1.0 m on a string that breaks when the tension exceeds 50 N. Calculate the maximum speed at which it can be whirled.

Accepted Answer

At the maximum, T = mv²/r (1) v = √(Tr/m) = √(50 × 1.0/0.25) (1) v = 14.1 m s⁻¹ (1)

Question 16

State the direction of the centripetal force on an object in circular motion, and describe what happens to the object if that force is suddenly removed.

Accepted Answer

The centripetal force acts towards the centre of the circle (1). If it is removed and no other force acts, then by Newton's first law the object travels in a straight line (1), along the tangent to the circle at the point of release (1).

Question 17

A pendulum bob of mass 0.40 kg swings on a string of length 1.8 m. At the lowest point of the swing its speed is 3.2 m s⁻¹ (g = 9.81 m s⁻²). Calculate the tension in the string at the lowest point.

Accepted Answer

At the lowest point the resultant upward force is T − mg (1) T − mg = mv²/r (1) T = 0.40 × 9.81 + (0.40 × 3.2²)/1.8 (1) T = 6.2 N (1)

Question 18

A laboratory centrifuge spins samples at 100 revolutions per second in a circle of radius 8.0 cm (g = 9.81 m s⁻²). Show that the centripetal acceleration of a sample is about 3200 times g.

Accepted Answer

ω = 2πf = 2π × 100 = 628 rad s⁻¹ (1) a = ω²r = 628² × 0.080 = 3.16 × 10⁴ m s⁻² (1) a/g = 3.16 × 10⁴/9.81 = 3219, which is about 3200 times g (1)

Question 19

Explain why the centripetal force on an object moving in a circle at constant speed does no work on the object.

Accepted Answer

The centripetal force acts towards the centre, at right angles to the velocity at every instant (1). The object therefore has no displacement in the direction of the force (1). Since W = Fs cos θ with θ = 90°, no work is done and the kinetic energy (and speed) stays constant (1).

Question 20

A particle moves at constant speed v round a circle of radius r. In a short time Δt its velocity turns through a small angle Δθ. Show, by drawing the two velocity vectors from a common point, that the change in velocity has magnitude Δv = vΔθ, and hence derive a = v²/r. Explain how the same triangle shows that the acceleration is directed towards the centre of the circle.

Accepted Answer

The speed is constant, so the two velocity vectors have the same magnitude v; each is a tangent, so while the radius turns through Δθ the velocity turns through the same angle Δθ (1)
Drawn from a common point, the change in velocity Δv is the third side of an isosceles triangle whose equal sides are v with Δθ between them; for a small Δθ that side is an arc of a circle of radius v, so Δv = vΔθ (1)
The time taken to sweep Δθ is Δt = Δθ/ω, from the definition of angular speed (1)
a = Δv/Δt = vΔθ/(Δθ/ω) = vω, in which Δθ cancels, so the result does not depend on the size of the step taken; substituting ω = v/r gives a = v²/r (1)
The two velocities lie symmetrically about the mid-point of the arc, so their difference lies along the radius at that point and is directed inwards, along neither of the two velocities (1)
As Δθ → 0 that mid-point closes on the position of the particle itself, so at any instant the acceleration is directed towards the centre of the circle (1)