Circular Motion

Question 1

What is Uniform circular motion

Answer 1

Motion along a circular path in which there is no change in speed, only in direction.

Question 1

In Uniform circular motion what is the direction of constant velocity and constant force?

Answer 1

Constant velocity is tangent to the path

Constant force is towards the center

Question 2

What happens when the central force is removed in uniform circular motion?

Answer 2

The ball will continue moving in a straight line. Centripetal force is needed to change direction

Question 3

Is there an outward force acting on a ball in uniform circular motion?

Answer 3

No, because the ball is moving at a tangent to the path. When the force is removed, it will continue in a straight line not outwards

Question 4

What are the key conceprts of acceleration in uniform circular motion?

Answer 4

• Speed is constant, but direction is always changing.
• Velocity is a vector and can be impacted by magnitude or direction or both.

-If the direction is changing, therefore velocity is changing so there is acceleration.

-If acceleration is present, there must be a force acting on the object (Newton’s 1/2 law)

-The direction of the force is always towards the center there for the direction of acceleration is also towards the center

Question 5

What is centripetal force?

Answer 5

A force acting on an object moving in a circular path which is directed toward the centre around which the object moves.

Question 6

What is the equation for linear and angular speed during uniform circular motion?

Answer 6

Linear speed, v = d/t = s/t (ms-1)

Angular speed, w = angle/time = 0/t (rad/s)

If the object has completed a full cycle (distance = circumference) in time T s, then the time T is known as Time Period.

Linear speed, v = distance/time = 2𝜋𝑟/T

Angular speed, w = angle/ time = 2𝜋/𝑇

𝑅𝑒𝑝𝑙𝑎𝑐𝑖𝑛𝑔 2𝜋/𝑇 𝑖𝑛 𝑙𝑖𝑛𝑒𝑎𝑟 𝑠𝑝𝑒𝑒𝑑 𝑓𝑜𝑟𝑚𝑢𝑙𝑎, 𝑣=𝑟ω

Question 7

A particle moves round the circumference of a circle with radius 10𝑚 at a speed of 20𝑚𝑠^(−1).
Calculate its angular speed.

Answer 7

v = rw

w = v/r = 20/10

=2

Question 8

How do you resolve x and y components?

Answer 8

ysin𝜃
xcos⁡𝜃

Question 9

What is the accelearation of an object moving on a circular path with constant angular speed?

Answer 9

r𝜔^2 directed towards the centre of the circle.

The position vector of P at time 𝑡 𝒔 =rcos𝜃 and rsin𝜃

At constant speed: 𝑑𝜃/𝑑𝑡=𝜔

Let 𝑡=0 when P is at A ⇒ 𝜃=𝜔𝑡

⇒𝒔=𝑟cos⁡𝜔𝑡 and 𝑟𝑠𝑖𝑛𝜔𝑡

differentiating s gives v =(−𝑟𝜔 sin⁡𝜔𝑡 and 𝑟𝜔 𝑐𝑜𝑠 𝜔𝑡)

differentiating v gives a=

𝒂=(−𝑟𝜔^2 cos⁡𝜔𝑡 and −𝑟𝜔^2 𝑠𝑖𝑛 𝜔𝑡) = −𝜔^2 (𝑟 cos⁡𝜔𝑡 and 𝑟 𝑠𝑖𝑛 𝜔𝑡)=−𝜔^2 𝒓

Question 10

What is another way of calculating acceleration in circular motion?

Answer 10

𝑎=𝑟𝜔^2 = 𝑣^2/𝑟

We know that 𝑣=𝑟𝜔⇒ 𝜔=𝑣/𝑟
Also 𝑎=𝑟𝜔^2
Substituting for 𝜔 gives: 𝑎=𝑟(𝑣/𝑟)^2=𝑣^2/𝑟

Question 11

A particle is moving on a horizontal circular path of radius 30cm with a constant angular speed of 4𝑟𝑎𝑑 𝑠^(−1). Calculate the acceleration of the particle.

Answer 11

𝑎=𝑟𝜔^2=0.3×4^2=4.8𝑚𝑠^(−2)

Question 12

What is the equation for acceleration towards the centre (centripetal acceleration)?

Answer 12

ac= v^2/R

Fc = mac

= mv^2/R

Question 13

A 3-kg rock swings in a circle of radius 5 m. If its constant speed is 8 m/s, what is the centripetal acceleration?

Answer 13

ac = 12.8 m/s
Fc = 38.4 N

Question 14

A skater moves with 15 m/s in a circle of radius 30 m. The ice exerts a central force of 450 N. What is the mass of the skater?

Answer 14

m = 60

Question 15

What is the nature of centripetal force Fc?

Answer 15

The centripetal force Fc is that of static friction fs

Question 16

How do you calculate the maximum speed for negotiating a turn without slipping

Answer 16

Fc = Fs

Fc = mv^2/R

fs = μs x mg

v = √μsgR
Where v is maximum speed without slipping

Question 17

How can optimum banking angle be calculated?

Answer 17

By resolving the vectors using trig

Nsin0 = mv^2/R

Ncos0 = mg

Tan 0 = v^2/gR

Question 18

What is does Newton’s second law state in relation to circular motion?

Answer 18

N2L states that all accelerating objects require a net or resultant force.

F = ma

Question 19

What direction are centripetal force and acceleration acting in?

Answer 19

Towards the center of the circle

Question 20

Is there work being done during uniform circular motion?

Answer 20

No, Since the net force is perpendicular to the direction of motion and work done = Force x distance, no work is being done on the force. This is also why there is no change in speed.

Question 21

What factors will cause an increase in the resultant centripetal force?

Answer 21

• The object rotates faster ( Has a larger angular velocity.)
• If the object has more mass
• If the object is further from the center of the circle