Applied Circular Motion Flashcards
A spin-dryer removes excess water from clothing by rotating a perforated drum at high speed.
The water is thrown out through the holes.
But if the centripetal force always acts towards the centre of the circle, why is the water thrown out?
The clothes keep moving in a circle because the contact force of the drum provides the centripetal force. But the water droplets can pass through the holes in the drum. The drum then exerts no force on them. So the water droplets obey Newton’s first law. They carry on moving in a straight line.
An object of mass 6.0 kg is moved in a vertical circle of radius 1.5 m at a constant speed of 5 m s-1. Calculate the maximum and minimum tensions in the string.
Top = 41N Bottom = 160N
The diagram shows a disc of diameter 120 mm that can turn about an axis through its centre.
The disc is turned through an angle of 30° in 20 ms. What is the average speed of a point on the edge of the disc during this time?
0.5π ms–1
π ms–1
1.5π ms–1
2π ms–1
xxx
A 0.10kg mass is to be placed on a horizontal turntable that is then rotated at a fixed rate of 78 revolutions per minute. The mass may be placed on the table at any distance, r, from the axis of rotation, as shown in Figure 2.
If the maximum frictional force between the mass and the turntable is 0.50 N, calculate the maximum value of the distance r at which the mass would stay on the turntable at this rate of rotation.
7.5 x 10^-2 m
An elastic cord has an unextended length of 13.0 cm.
One end is attached to a fixed point P.
A mass of 5.0 N is attached to the other end. The cord
extends to a length of 14.8 cm.
The cord and mass are made to rotate at a constant
angular velocity of ω in a vertical plane about point P.
When the cord is vertical and above P, its length is the
unextended length of 13.0 cm.
Show that the angular velocity ω of the cord and mass is 8.7 rad s-1
Calculate the length L of the cord when it is vertically below P as shown in the diagram, assuming the cord obeys Hooke’s law
- 7 rad s ^-1
17. 2cm
An electron is moving at 1% of the speed of light in a circle of radius 1.4 m in a particle accelerator. Calculate the force the magnetic field is exerting on the electron, and state the direction of the force.
5.86 x 10⁻¹⁸ N
A single stationary particle decays into a proton and an antiproton, which move in curved tracks of radius 5.06 cm away from each other. If the force on each particle is 5 x 10-10 N, calculate the speed each particle is moving at.
1.23 x 10⁸ ms⁻¹
An electron is accelerated to a velocity of 2.1 x 107 ms-1 by a particle accelerator, in a magnetic field of strength B = 0.55 mT. Calculate the radius of the electron’s path. Use the equation F = Bqv to calculate the force on the electron due to the field.
0.217 m
