Centripetal Force Flashcards
A rider on a merry-go-round is 3.5 m from the centre point. They are travelling with an instantaneous velocity of 4.8 ms-1.
Calculate their angular displacement in one second
Calculate their angular velocity
Calculate the time period of a full circuit
Calculate the frequency of the merry-go-round
Calculate the centripetal acceleration acting on the rider
1.37 rad
ω = 1.37 rad s-1
T = 4.58 s
f = 0.18 Hz
a = 6.58 ms-2
Fill in the blanks
The linear velocity of the moving object is constantly ________.
Newton’s first and second laws tell us that the object must be __________ if its velocity is ____ _______.
Newton’s first and second laws tell us that the object must have a ________ force in the ______ direction to cause acceleration
The resultant force must act ______ the centre
The linear velocity of the moving object is constantly changing
Newton’s first and second laws tell us that the object must be accelerating if its velocity is not constant
Newton’s first and second laws tell us that the object must have a resultant force in the same direction to cause acceleration
The resultant force must act towards the centre
True or False?
Objects travelling in a circle must experience a resultant force acting towards the centre of the circle.
True
There may or may not be other forces acting on the object, but the centripetal force must be the resultant force
Is centripetal a type of force?
No
What direction does centripetal force act in?
Centripetal acceleration is always directed towards the centre of the circle that the object is moving in
What is the centripetal force needed for a car mass 1000kg to follow a curve of radius 50m at 18ms^-1 (≈40mph)
Mass: 1000 kg Velocity: 18.0 m s-1 Radius: 50.0 m Force, F = (mass x velocity2) ÷ radius = (1000 x 18.02) ÷ 50.0 = 6480 N
A 44.0 kg child is standing on the outer edge of a playground roundabout with an angular velocity of 1.40 rad s-1. Calculate the centripetal force required to maintain their position 2.30 m away from the centre.
F = m r ω^2 = 44 x 2.3 x 1.42 = 198 N
A 2.7 kg lasso is spun at a constant angular velocity of 3.6 rad s-1. What is the tension in the rope (centripetal force) at the end, 1.2 m from your hand?
F = m r ω^2 = 2.7 x 1.2 x 3.62 = 42 N
A 0.8 kg ball is attached to a string that is 1.2 m long. It is spun so that it completes two full rotations every second. What is the centripetal force acting on the ball?
F = m r ω^2 = 0.80 x 1.2 x (4π)2 = 150 N
A 0.78 kg rock tied to a string is moving at a constant speed of 8.0 ms-1 in a circle of radius 7.0 m. What is the centripetal force acting on the rock?
F = (m v^2) / r = (0.78 x 8.02) / 7.0 = 7.1 N
A 35 g ball on the end of a rope is moving at a constant speed of 9.7 ms-1 in a circle of radius 1.6 m. What is the centripetal force acting on the rock?
F = (m v^2) / r = (0.035 x 9.72) / 1.6 = 2.1 N
A 1200 kg car passing the top of a hill with radius of curvature 150 m at a speed of 16 ms-1. What centripetal force is provided by the friction with the road?
F = (m v^2) / r = (1200 x 162) / 150 = 2000 N
A 25 kg child is riding a merry-go-round with a radius of 5 m. What is the centripetal force on the boy if the merry-go-round is travelling at 0.8π rad s^-1
790 N
A car of mass m is driving around a circular track of radius r at a constant velocity of v. The centripetal force, provided by friction, acting on the car is F. If the car’s velocity is doubled, what is the new centripetal force required for the car to drive on the circular track?
4F
During an athletic hammer throw, a hammer is moving at a constant angular velocity of 25.00 ms-1 in its circular orbit. Being a regulation hammer, it weighs 7.257 kg, and the length of its rope is 121.5 cm. What is the magnitude of the tension exerted by the athlete along the hammer’s rope?
3733 N
