Evans SPH 4U

Physics Grade 12

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 10 ^{2
}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 10 ^{1} m/s [W]**

d) The centripetal acceleration at B.
**Answer: 3.0 x 10 ^{1} 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

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 10 ^{2}
N [down] BOTTOM: 2.0 x 10^{2} N [Up]**