5 - Circular Motion

Centripetal/Radial Acceleration (a_c/a_r)

acceleration directed towards the center of a circular path

• changing direction

• constant speed

• velocity tangent to path @ that instant

• points to center, comes from circular motion

Calculation of a_r

$$a_c = \frac{v^2}{r}$$

• $v$ →tangential speed

• $r$ → radius of the circular path

What do you need to know to calculate a_r?

• need to know the speed & size of path

• tight curve = fast & short time

• long curve = slow & more time/safer

What happens if the curve is the same but speed is doubled?

2x speed = ½ time = 4x accel, 3x speed = 1/3 time = 9x accel

• find accel = inc velocity & dec time

• accel depends on v²

What happens if the radius of the curve is doubled?

same v, t doubles to go around the bigger circle

• a_r= v/2t = v²/r

• a_r depends on 1/r → inc r = dec t

Period (T)

amount of time (s) to make 1 rotation/revolution

• applicable to any object going through circular motion

• object going around circle@ v → distance = speed x time

• T = $\frac{\left(2\pi r\right)}{v}$

• describe in revolutions/time = rpm (T= 1/rpm)

Analyzing Circular Motion

to have an a_r = need a net force pointing in that direction

• point 1 axis towards center of circle

• accel becomes a_r

• F_net becomes centripetal force

• sum Fr = ma_r= mv²/r

A car is traveling around a curve at a steady 45 mph. Is the car accelerating?

Yes, the car is accelerating due to the change in direction, even though its speed remains constant.

A car is traveling around a curve at a steady 45 mph. What direction is the car’s acceleration in?

towards the center of the curve

A small piece of ceramic flies off the rim of a circular potter's wheel that is spinning fast. In which direction will the ceramic piece travel the moment it leaves the wheel?

Along a straight tangent to the circular rim at the point where the ceramic left the wheel        

If the speed of an object in circular motion is doubled, what should be the new radius if the radial acceleration is to remain unchanged?    

The new radius should be quadrupled to keep the radial acceleration constant.

• ar is directly proportional to v²

• ar is inversely proportional to radius

• speed soubled = square of speed = 4x square og speed

You see a penny sitting on a turntable rotating with constant speed. How many forces are acting on the penny?

The downward force of the earth on the penny, the upward force of the turntable on the penny, and the inward friction force of the turntable on the penny

What is the direction of the acceleration of an object in uniform (constant speed) circular motion?    

towards center of the circle

If an object is moving with uniform circular motion at a constant speed v and radius r, then its tangential acceleration and radial acceleration are _______ and _______, respectively.    

zero, v²/r

Why is it difficult for a high-speed car to negotiate an unbanked turn?

The magnitude of the friction force might not be enough to provide the necessary radial acceleration.

If you put a penny on the center of a rotating turntable, it does not slip. However, if you place the penny near the edge, it is likely to slip off. Which answer below explains this observation?

The radial acceleration is greater at the edge and the friction force is not enough to keep the penny in place.

An object moves in a circular path at a constant speed. What is the direction of the sum of the forces exerted on the object?

The sum of the forces is directed toward the center of the circular path.

Two objects attract each other gravitationally. If the mass of each object doubles, how does the magnitude of the gravitational force that they exert on each other change?

The gravitational force increases by a factor of 4.

A hypothetical planet has a mass one-third of and a radius three times that of Earth. What is the acceleration due to gravity on the planet in terms of g, the acceleration due to gravity on Earth?

The acceleration due to gravity on the hypothetical planet is given by the formula $g' = \frac{G \cdot m}{r^2}$, where the mass of the planet is one-third of Earth's mass and the radius is three times that of Earth. Thus, the calculation yields an acceleration due to gravity of $g' = \frac{g/3}{(3)^2} = \frac{g}{27}$.

Why does a satellite in a circular orbit travel at a constant speed?

There is no component of force exerted along the direction of motion of the satellite.

A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy is closer to the center. Who has greater tangential acceleration?

Both the girl and the boy have zero tangential acceleration.