Circular Motion

1) A CD spins in the player with a frequency of 1800 revolutions per minute. What is the period of the spin of the disc?

Period = 1/frequency = .0006 seconds (the time for one revolution)

2) Luigi twirls a round piece of pizza dough overhead with a frequency of 45 revolutions per minute.

a) find the linear speed of a stray piece of pepperoni stuck on the dough 10 cm from the pizza's center.

v=2r/T = 2(3.14)10 cm / .022 sec = 28.5 m/s

remember from #1 that period(time) = 1/freq

b) Describe the motion of the pepperoni if it flies off the pizza while the pizza is spinning.

it flies of tangent to the pizza in a straight line

3) Gravitational fields with strength g can be simulated by g = v2/r

A 2.0 km radius space station needs to be rotated to simulate earth's gravitational field. How fast must it be rotated?

9.8 m/s2 = v2/2000 m , v2 = 9.8 m/s2 x 2000m = 19600 m2/s2, so v = 140 m/s

4) A tennis ball is being twirled on a string. If the mass of the ball is tripled by what factor does the

tension in the string increase? __Three____ by what factor does the centripetal force increase? __Three____

If the velocity of the ball is doubled, by what factor does the centripetal force change? ___Four_____

If the radius of the string is doubled, by what factor does the centripetal force change? ___Half______

remember that Fc= mv2/r

5) A 1500 kg car moving on a flat road negotiates a curve that has a radius of 35 m. If the frictional force acting on the car is 7500 N, what is the maximum speed the car can have in order to make the turn successfully?

Fc = mv2/r , and the road is pushing the car into the curve, so 7500 N = (1500 kg)(v)2/35 m ,

so v2 = 175 m2/s2 and v = 13.2 m/s

 

CENTRIPETAL MOTION

Fc = mv2/r

1.) A 150-g ball at the end of a string is swinging horizontally in a circle of radius 1.15 m. The ball makes exactly 2.00 revolutions in a second. What is its centripetal force?

Period = 1/2.00 = .5 sec , v = 2r/T = 14.5 m/s

Fc = .150 kg(14.5 m/s)2 / 1.15 m = 27.2 N

2.) The moon's nearly circular orbit about the earth has a radius of about 385,000 km and a period of 27.3 days. Determine the force needed to keep the moon in orbit.

v = 2()(385000000m)/27.3 = 88600000 m/s

Fc = 7.35 x 10 22 kg(88600000m/s)2/385000000 = 1.5 x 10 30 N

3.) What is the force required to keep a car moving in a circle around a curve if the 1,000 kg car rounds the curve on a flat road of radius 50 m at a speed of 14 m/s?

Fc = 1000kg(14 m/s)2/50m = 3920 N

4)What is the centripetal force needed to keep a 3.0 kg object moving in a circle of 2.0 m radius at a speed of 4.0 m/s?

Fc = 3.0 kg (4.0 m/s)2/2.0 m = 24 N

5) A force of 26 N is applied to a 0.6 kg stone to keep it rotating in a horizontal circle of radius 0.4 m. What is its velocity?

26 N = 0.6 kg (v)2 / 0.4 m, so v = 4.16 m/s

Can a small force exert a greater torque than a larger force? Explain.

Yes, if the smaller force is exerted on a longer lever arm

Torques

  1. A force of 10 N is applied at the end of a wrench handle that is 30 cm long. The force applied is in a direction perpendicular to the handle in order to turn a nut at the opposite end of the wrench.
    1. What is the torque applied to the nut by the wrench?

    Torque = force x lever arm, so Torque = 10 N x .3 m = 3 Nm

    1. What would the torque be it the force were applied halfway up the handle instead of at the end?

    Torque = 10 N x .15 m = 1.5 Nm

  2. A weight of 15 N is located at a distance of 20 cm from the fulcrum of a simple balance beam. At what distance should a weight of 20 N be placed on the opposite side in order to balance the system?

Balanced beam so torque on one side equals torque on the other, so 15 N x .2 m = 20 N x LA, so LA = .6 m

  1. A weight of 5 N is located 10 cm from the fulcrum of a simple balance beam. What weight should be placed at a point 4 cm from the fulcrum on the opposite side in order to balance the system?

5 N x .1 m = F x .04 m , so F = 12.5 N