Evans SPH 4U

Unit 3: Circular Motion

Note 1: Centripetal Acceleration

For derivation of formulas refer to: http://dev.physicslab.org/Document.aspx?doctype=3&filename=CircularMotion_CentripetalAcceleration.xml

Reference Chapter 3.1

~ Circular motion with weight on string - if let go it continues in tangential (not circular) motion (Law of Inertia).

~ Swing a bucket of water in a circle. What causes the water to stay in? Water wants to continue in a straight line (law of inertia) but you are providing a centripetal force to pull it back in.  From the bucket's frame of reference there is a force pushing out.

Uniform Circular Motion: Motion with constant speed and radius.

Centripetal Acceleration: Instantaneous acceleration directed toward the centre of the circle.

Centripetal Force: Force that causes centripetal acceleration since Newton's 2nd law states that acceleration is caused by a net force.  Centripetal force causes a change in the direction of the velocity rather than a change in magnitude.

Centrifugal Force: Fictitious force in an accelerating frame of reference. i.e. It pushes the object to the outside of the circle.

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Example 1:

Find the magnitude and direction of the centripetal acceleration of a piece of lettuce on the inside of a rotating salad spinner.  The spinner has a diameter of 19.4 cm and is rotating at 780 rpm.  The rotation is clockwise as viewed from above.  At the instant, the lettuce is moving east (use your units!) Answer: 6.5 x 102 m/s [into centre]

Example 2:

A rubber ball is attached to a string 0.75m long and whirled around clockwise in a horizontal circle.  (You need to refer to the diagram you copied from the overhead in class or see me for the diagram). The ball takes 0.25 s to move from A to B. Find:

a) The velocity at A Answer: 4.7 m/s [SE]

b) The average velocity from A to B. Answer: 4.2 m/s [S]

c) The average acceleration from A to B Answer: 2.7 x 101 m/s [W]

d) The centripetal acceleration at B. Answer: 3.0 x  101 m/s [NW]

~ Vertical circular motion ~ students should experience greater force required at bottom of loop.

Example 3:

A stone of mass of 1.0kg is tied to a 2.0m long string.  The stone is swung in a vertical circle with a constant speed of 5.0 m/s.  Calculate the tension in the string when it is:

a) At its highest point in the circular path Answer: 2.7 N [Down]

b) At its lowest point in the circular path Answer: 22.3 N [Up]

c) At what speed is the object going to feel no tension at the top?  Answer: 4.4 m/s At this speed what will the tension be at the bottom? Answer: 19.6 N [Up] ( it works out to be 2mg)

d) If the mass is increased to 10kg at what speed is the object going to feel no tension at the top? Answer: Still at 4.4 m/s since mass cancels out (it is mass independent)

Example 4:

A car is moving at uniform velocity in a quarter circular turn.  The back door is faulty and will open if the force on the door is >20.0N.  There's a goldfish bowl on the back car seat with a mass of 1.0kg.  The turn in the road has a radius of 30.0 m. What is the maximum speed the car can take the turn without losing the goldfish on the road? Answer: 24.5 m/s

Example 5:

A 3.5 kg steel ball is swung at a constant speed in a vertical circle of radius 1.2m on the end of a light, rigid steel rod.  If the ball has  a frequency of 1.0 Hz, calculate the tension in the rod due to the mass at the top and at the bottom positions.

Answer: TOP: 1.3 x 102 N [down]     BOTTOM: 2.0 x 102 N [Up]