6 Motion in a circle

Question 1

Define angular displacement.

Answer 1

Angular displacement is defined as the angle an object makes with respect to a reference line.

Question 2

Define radian.

Answer 2

The angle subtended by an arc length equal to the radius of the arc.

Question 3

Define an angle in radians using s and r.

Answer 3

Angle theta = s/r where s is the arc length and r is the radius of the circular.

Question 4

Define angular velocity. State its S.I. unit and whether it is a scalar or vector quantity. State the equations used to calculate angular velocity.

Answer 4

Represented by omega, angular velocity is defined as the rate of change of its angular displacement with respect to time. Omega = dtheta/dt = 2pi/T = 2pi*f. The vector quantity which has both a magnitude and direction (clockwise/anti-clockwise) is measured by rads^-1.

Question 5

Define period.

Answer 5

The period of an object in circular motion is the time taken for it to make one complete revolution.

Question 6

Define frequency.

Answer 6

The frequency of an object in circular motion is the number of revolutions made per unit time. (Note that it need not be complete revolutions)

Question 7

State the formula relating frequency and period.

Answer 7

f = 1/T

Question 8

State the relationship between angular velocity and linear velocity/tangential velocity.

Answer 8

s = rtheta
ds/dt = rdtheta/dt
v = r*omega

Question 9

State the relationship between period and linear speed.

Answer 9

v = Circumference of circle/T
T = 2rpi/v

Question 10

Explain why an object must always experience a force when moving in a uniform circular motion.

Answer 10

According to N1L, the object will continue to move at its constant speed in the same direction unless there is a resultant force acting on it. Hence, there must be a force acting on the object to constantly change its direction of motion.

Question 11

Explain why the force (acceleration) is always directed towards the centre of the circle/the force (acceleration) is always perpendicular to the motion of the object. State the name of the acceleration and force in a circular motion.

Answer 11

As the object is moving at a constant speed, there must be no component of force (acceleration) in the direction of the motion of the object. Therefore, the direction of this force has to be perpendicular to the direction of motion. Using a vector diagram, it can be found that the direction of the change in velocity (acceleration) is towards the centre of the circle. ??? The acceleration of this nature is called centripetal acceleration and the force is called the centripetal force.

Question 12

State the formula to calculate centripetal acceleration and centripetal force.

Answer 12

a = v^2/r = v*omega = r*omega^2
F = ma = mv^2/r = mv*omega = mr*omega^2

Question 13

Explain if there is work done by centripetal force.

Answer 13

Although a centripetal force is needed to keep an object moving in a circle, the force does no work don’t the object since there is no displacement of the object in the direction of the force.

Question 14

Explain if centripetal force is a new force.

Answer 14

Centripetal force is not a new force and should not be included in the free-body diagram of an object moving in a uniform circular motion. It can be the tension in a string, gravitational force of attraction or frictional force between two bodies or a combination of forces. If the force that produces the centripetal acceleration is removed, the object does not continue to move in a circular path and will move tangentially to its original path due to inertia (move in the direction of tangential velocity). This because centripetal force is the resultant of all the other forces acting on the object.

Question 15

Explain why a cyclist has to lean inwards towards the centre of the circular motion.

Answer 15

The centripetal force necessary for the cyclist to execute the turn is provided by the frictional force acting on the bicycle. This force produces a clockwise moment about the centre of gravity which tends to turn the cyclist outwards. When the cyclist leans inward, the normal contact force produces a counterbalance moment about the centre of gravity so that no rotation would occur.

Question 16

Explain why a car may skid off a track when it is travelling on a circular path.

Answer 16

As a car travels faster, the centripetal acceleration required to maintain the same radius increases since centripetal acceleration (provided by friction) = v^2/r. Therefore, the sideward friction required to keep the car on the circular path must increase. When the force required to keep the car on the circular path is greater than the maximum sideward frictional force that can be provided by the track, the car will skid off the track - radius increases.

Question 17

Explain how banking helps to increase safety on roads and cut down the wear and tear of tyres and of road surfaces.

Answer 17

Banking involves road surfaces being tilted at sharp bends to minimise friction. It helps to provide part of the centripetal force from the normal contact force and thus reduce the sideward friction that is needed to produce the centripetal attraction.

Question 18

State the equation used to find the ideal velocity for a car travelling round a circular bend along a banked track. Explain what happens when the actual velocity of a car is more than the ideal velocity.

Answer 18

Vertically: Ncos theta = mg
Horizontally = Nsin theta = mv^2/r

tan theta = v^2/rg

This angle of banking, theta is ideal as no frictional forces are needed to provide for centripetal acceleration. Only the horizontal component of the normal contact force is required and provides the centripetal force required to keep the car moving round the bend.

tan theta = Ac/g = videal^2/r
where Ac is the centripetal acceleration

When the actual velocity of a car is more than the ideal velocity, sideward frictional force on the tyres will act down the slope as the car tends to slide up the bank. When the actual velocity of a car is less than the ideal velocity, sideward frictional force on the tyres will act up the slope as the car tends to slide down the bank.

Question 19

Explain why a body undergoing non-uniform circular motion has varying tangential speed.

Answer 19

Due to the non-zero tangential acceleration in addition to centripetal acceleration. In a uniform circular motion, the net force is always directed towards the centre of the circular motion. This is not true in a uniform circular motion where the tangential speed is constantly changing. CLARIFY AFTER CTS

Question 20

Explain why a train is more likely to lose contact with the track at the top of its circular motion. Explain how to find the minimum speed needed for the train and the minimum speed at which the train must enter the loop to remain on track throughout the motion around a vertical circular loop.

Answer 20

Pg 14 and 15 of notes.

Question 21

Which variable represents the rate of rotation?

Answer 21

Omega = 2pi/T

Question 22

State the conditions required for a particle to perform uniform circular motion.

Answer 22

A resultant force acting on the particle is always perpendicular to the direction of its linear speed/velocity and is directed towards a fixed point. The resultant force must be of constant magnitude.

Question 23

Explain from both the dynamic and work-energy points of view why the speed of a particle that is in uniform circular motion is constant.

Answer 23

From the dynamical point of view, since the resultant (or centripetal) acceleration is perpendicular to the instantaneous velocity, the magnitude of the velocity will be unchanged hence the linear speed of such particle remains constant.

From the work-energy point of view, as the resultant force acts perpendicularly to the direction of motion, work done by the resultant force is zero. This does not cause a change to the kinetic energy of the particle and hence its speed remains constant.

Question 24

Explain why the velocity and acceleration of a particle in uniform circular motion are both not constant.

Answer 24

Both the velocity and acceleration are not constant (changing continuously) as the direction of motion and the direction of the acceleration is constantly changing. (Note that for uniform circular motion, while the magnitudes of the velocity and acceleration are constant, their directions are not. They are both vector quantities, and hence direction must be considered.)

Question 25

What is the relationship between sin theta and theta for small angles?

Answer 25

sin theta approximately equals theta

Question 26

Sketch the cross-section of the aircraft how a plumbline is orientated with respect to the aircraft.

Answer 26

Tutorial D4e.