Basic Physics

# Nonuniform circular motion – problems and solutions

1. A wheel 1 meter in radius accelerates uniformly at 2 rad/s2. Determine the angular acceleration and the angular speed of the wheel, 2 seconds later.

Known :

Angular acceleration (α) = 2 rad/s2

Wanted: angular acceleration and angular speed after 2 seconds.

Solution :

(a) Angular acceleration in 2 seconds

Angular acceleration is constant, thus after 2 seconds, angular acceleration of the wheel is 2 rad/s2.

(b) Angular speed in 2 seconds

Angular acceleration 2 rad/s2 means the angular speed increases 2 radians/second each 1 second. After 1 second, angular speed = 2 radians/second. After 2 seconds, angular speed = 4 radians/second.

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2. A particle accelerates uniformly from rest to 60 rpm in 10 seconds. Determine the magnitude of angular acceleration!

Known :

The initial angular velocity (ωo) = 0

The final angular velocity (ωt) = 60 rpm = 60 revolutions / 60 seconds = 1 revolution / second = 6,28 radians/second

Time interval (t) = 10 seconds

Wanted : Angular acceleration (α)

Solution : ωo = the initial angular velocity, ωt = the final angular velocity, α = the angular acceleration, t = time interval, θ = angle.

ωt = ωo + α t

6.28 = 0 + α (10)

6.28 = 10 α

α = 6.28 / 10

The magnitude of the angular acceleration = 0.628 rad/s2

3. An object slows down from 20 rad/s to 10 rad/s in 4 seconds. Determine the magnitude of angular acceleration!

Known :

Time interval (t) = 4 seconds

The initial angular velocity (ωo ) = 20 rad/s

The final angular velocity (ωt) = 10 rad/s

Wanted : the magnitude of the angular acceleration (α)

Solution :

ωt = ωo + α t

10 = 20 + α (4)

10 – 20 = 4 α

-10 = 4 α

α = -10 / 4

The magnitude of the angular acceleration is -2.5 rad/s2. Negative sign means the object is decelerating. Acceleration = the angular speed increases, deceleration = the angular speed decreases.

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4. An object is accelerated for 2 seconds from 10 rad/s to 2 rad/s2. Determine angle rounded by the object!

Known :

the initial angular velocity (ωo ) = 10 rad/s

the angular acceleration (α) = 2 rad/s2

time interval (t) = 2 seconds

Wanted : angle (θ)

Solution :

θ = ωo + ½ α t2

θ = (10)(2) + ½ (2)(22)

θ = 20 + (1)(4) = 20 + 4

5. A car’s wheel slows down from 20 rad/s to rest after round 20 radians. Determine the magnitude of the angular acceleration of the wheel!

Known :

the initial angular speed (ωo) = 20 rad/s

the final angular speed (ωt) = 0

Wanted : the magnitude of the angular acceleration (α)

Solution :

ωt2 = ωo2 + 2 α θ

0 = 202 + 2 α (20)

0 = 400 + 40 α

400 = – 40 α

α = – 400 / 40

6. A rod PQ with length of 60 cm rotated about point Q as the axis of rotation and PQ as the radius of circle. The rod PQ accelerated from rest to 0.3 rad/s2. What is the linear speed of point P at t = 10 seconds, if the angular initial position is 0.

Known :

Length of rod PQ = radius of circle (r) = 60 cm = 60/100 m = 0.60 m

The initial angular speed (ωo) = 0 rad/s

Angular acceleration (α) = 0.3 rad s-2

The initial angular position (θo) = 0

Wanted : Linear speed (v) of point P at t = 10 seconds

Solution :

The final angular speed after 10 seconds :

ωt = ωo + α t = 0 rad/s + (0.3 rad s-2)(10 s) = 3 rad/s

The final linear speed after 10 seconds :

v = r ω = (0.6 m)(3 rad/s) = 1.8 m/s

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7. An object rotates with the initial speed of 4 rad/s and the angular acceleration is 0.5 rad/s2. What is the speed of object after 4 seconds.

Known :

The initial angular speed (ωo) = 4 rad/s

Angular acceleration (α) = 0.5 rad/s2

Time interval (t) = 4 seconds

Wanted : Object’s speed after 4 seconds (ωt)

Solution :

ωt = ωo + α t

ωt = 4 + (0.5)(4)

ωt = 4 + 2

8. A wall clock with diameter of 10 cm has three needles, each to show the hours, minutes and seconds. Comparison of the number of rounds of the hour needle: the minute needle: the second needle.

A. 1 : 3 : 180

B. 1 : 12 : 720

C. 4 : 12 : 180

D. 4 : 12 : 720

Known :

1 hour = 60 minutes

12 hours = (12)(60 minutes) = 720 minutes

Angular speed of the hour needle = 1 revolution / 12 hours = 1 revolution / 720 minutes

Angular speed of the minutes needle = 1 revolution / 1 hour = 1 revolution / 60 minutes

Angular speed of second needle = 1 revolution / 1 minute

Wanted: Comparison of the number of rounds of the hour needle: the minute needle: the second needle

Solution :

The equation of circular motion :

Angular speed = number of revolution / time interval

Number of revolution = angular speed x time interval

In the same time interval, for example, 1 minute, how many revolution of hour needle, minute needle, and the second needle.

Number of revolution of the hour needle = angular speed x time interval = (1 revolution / 720 minutes)(1 minute) = 1/720 revolutions

Number of revolution of the minute needle = angular speed x time interval = (1 revolution / 60 minutes)(1 minute) = 1/60 revolutions

Number of revolution of the second needle = angular speed x time interval = (1 revolution / 1 minute)(1 minute) = 1/1 revolution

Comparison of a number of revolutions :

Number of revolution of the hour needle: number of revolution of minute needle : number of revolution of the second needle.

1/720 : 1/60 : 1/1

1/720 : 12/720 : 720/720

1 : 12 : 720

9. A ball tied with a rope. The ball is rotated so that it moves in a circular plane parallel to the surface of the earth. In this motion, the ball accelerates because…..

A. Friction of air

B. Weight of ball

C. Tension force

Solution :

Newton’s second law of motion states that an object is accelerated if there is a resultant force. The ball is connected to the rope and when the rope rotated, the ball also rotates. When the ball rotates (the ball moves in a circle), the ball undergoes centripetal acceleration. All the moving objects are circular centripetal acceleration. Centripetal acceleration is caused by centripetal force. The centripetal force for this case is the tension force.