Circular Motion :- Example

Note: This is an example sheet with solutions and is not regenerating.

Circular Motion

Take "g", the acceleration due to gravity as 9.8 m/s/s unless told otherwise.

Q.1. Jack was driving a car of mass 800kg at a speed of 13 m/s around a circular track of radius 36m.
a. What is the centripetal force acting on the car? (Nearest whole number)
b. What is the acceleration of the car? (2 d.p)

Q.2. Lisa was driving a car of mass 1500 kg in a circular path of radius 39 m. The car exerts a frictional force between the tyres and the track of half its weight.
What is the maximum speed with which it can travel around the track? (1 d.p.)

Q.3. Dylan was driving a car of mass 800kg around a circular track of radius 49m. A centripetal force of magnitude 2000 newtons was acting on the car.
a. What is the speed of Dylan's car? (1 d.p.)
b. What is the acceleration of the car? (1 d.p.)

Q.4. Sophie was driving a car of mass 1500kg around a circular track at a speed of 43m/s. A centripetal force of magnitude 7500 newtons was acting on the car
What is the radius of the circular track? (1 d.p)

Q.5. The road at the top of a hill is in the shape of an arc of a vertical circle of radius 9 m.
What is the maximum speed with which Cate can drive her car of mass 1400 kg over the hill without losing contact with the road?

Q.6. The road at the bottom of a hill is concave up and is in the shape of an arc of a vertical circle of radius 13 m. Paul drives his car of mass 900 kg down the hill and up the other side at a speed of 3 m/s.
What is the reaction force of the road on the car at the bottom of the hill?

Q.7. Galileo wants to ride a motorbike around 'The Wall of Death' at the Easter Show and ride in a vertical circle of radius 11 m. The mass of Galileo and his motorbike is 150kg.
What speed must he maintain to ride in a vertical circle without falling? (2 d.p.)

Answers:

Q1:.....(a)3756 Newtons towards the centre. (b) 4.69 m/s/s towards the centre.
Q2:.....13.8 m/s
Q3:..... (a) 11.1m/s. (b) 2.5m/s/s towards the centre
Q4:..... (a) 369.8 m Q5:..... (a) 9.4m/s
Q6:..... (a) 9443 Newtons up Q7:..... (a) 10.38 m/s

Circular Motion : Answers

Take "g", the acceleration due to gravity as 9.8 m/s/s unless told otherwise.

Q.1. Jack was driving a car of mass 800kg at a speed of 13 m/s around a circular track of radius 36m.

a. What is the centripetal force acting on the car? (Nearest whole number)
Fc = mv2/r
Fc = (800 x 132)/36 = 3756N towards the centre (nearest whole number)

b. What is the acceleration of the car? (2 d.p)
ac = v2 /r   = 132 /36 = 4.69 ms-2 towards centre

Q.2. Lisa was driving a car of mass 1500 kg in a circular path of radius 39 m. The car exerts a frictional force between the tyres and the track of half its weight.
What is the maximum speed with which it can travel around the track? (1 d.p.)

Friction provides centripetal force so
Fc = mv2/r
mg/2 = mv2/r
g/2 = v2/r
v2 = rg/2 = (39 x 9.8) /2 = 191.1
v = 13.8m/s (1 d.p.)

Q.3. Dylan was driving a car of mass 800kg around a circular track of radius 49m. A centripetal force of magnitude 2000 newtons was acting on the car.
a. What is the speed of Dylan's car? (1 d.p.)

Fc = mv2/r
2000 = 800 x v2/49
v2 = (2000 x 49)/800 = 122.5
v = 11.1 m/s

b. What is the acceleration of the car? (1 d.p.)
ac = v2 /r = 122.5 /49 = 2.5 ms-2 towards centre

Q.4. Sophie was driving a car of mass 1500kg around a circular track at a speed of 43m/s. A centripetal force of magnitude 7500 newtons was acting on the car.
What is the radius of the circular track? (1 d.p)
Fc = mv2/r
7500 = 1500 x 432/r
r = 1500 x 432 / 7500 = 369.8m

Q.5. The road at the top of a hill is in the shape of an arc of a vertical circle of radius 9 m.
What is the maximum speed with which Cate can drive her car of mass 1400 kg over the hill without losing contact with the road?

Fc = mv2/r = mg
v2 = rg = 9 x 9.8 = 88.2 v = 9.39 m/s

Q.6. The road at the bottom of a hill is concave up and is in the shape of an arc of a vertical circle of radius 13 m. Paul drives his car of mass 900 kg down the hill and up the other side at a speed of 3 m/s.
What is the reaction force of the road on the car at the bottom of the hill?

Reaction force = mg + mv2/r
R = 900 x 9.8 + (900 x 32)/13 = 8820 + 623 = 9443 Newtons up

Q.7. Galileo wants to ride a motorbike around 'The Wall of Death' at the Easter Show and ride in a vertical circle of radius 11 m. The mass of Galileo and his motorbike is 150kg.
What speed must he maintain to ride in a vertical circle without falling? (2 d.p.)

Fc = mv2/r = mg v2 = rg = 11 x 9.8 = 107.8 v = 10.38 m/s

Answers:

Q1:.....(a)3756 Newtons towards the centre. (b) 4.69 m/s/s towards the centre.
Q2:.....13.8 m/s
Q3:..... (a) 11.1m/s. (b) 2.5m/s/s towards the centre
Q4:..... (a) 369.8
Q5:..... (a) 9.4m/s
Q6:..... (a) 9443 Newtons up
Q7:..... (a) 10.38 m/s