# Nonuniform circular motion – problems and solutions

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

__Known :__

Radius (r) = 1 meter

Angular acceleration (α) = 2 rad/s^{2}

__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/s^{2}.

(b) __Angular speed in 2 seconds__

Angular acceleration 2 rad/s^{2} 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.

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

__Known :__

The initial angular velocity (ω_{o}) = 0

The final angular velocity (ω_{t}) = 60 rpm = 60 revolutions / 60 seconds = 1 revolution / second = 6,28 radians/second

Time interval (t) = 10 seconds

__Wanted :__ Angular acceleration (α)

__Solution :__

ω_{o} = the initial angular velocity, ω_{t} = the final angular velocity, α = the angular acceleration, t = time interval, θ = angle.

ω_{t} = ω_{o} + α t

6.28 = 0 + α (10)

6.28 = 10 α

α = 6.28 / 10

α = 0.628 rad/s^{2}

The magnitude of the angular acceleration = 0.628 rad/s^{2}

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 (ω_{o }) = 20 rad/s

The final angular velocity (ω_{t}) = 10 rad/s

__Wanted__ : the magnitude of the angular acceleration (α)

__Solution :__

ω_{t} = ω_{o }+ α t

10 = 20 + α (4)

10 – 20 = 4 α

-10 = 4 α

α = -10 / 4

α = – 2.5 rad/s^{2}

The magnitude of the angular acceleration is -2.5 rad/s^{2}. Negative sign means the object is decelerating. Acceleration = the angular speed increases, deceleration = the angular speed decreases.

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

__Known :__

the initial angular velocity (ω_{o }) = 10 rad/s

the angular acceleration (α) = 2 rad/s^{2}

time interval (t) = 2 seconds

__Wanted :__ angle (θ)

__Solution :__

θ = ω_{o }+ ½ α t^{2}

θ = (10)(2) + ½ (2)(2^{2})

θ = 20 + (1)(4) = 20 + 4

θ = 24 radians

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 (ω_{o}) = 20 rad/s

the final angular speed (ω_{t}) = 0

Angle (θ) = 20 radians

__Wanted :__ the magnitude of the angular acceleration (α)

__Solution :__

ω_{t}^{2} = ω_{o}^{2} + 2 α θ

0 = 20^{2} + 2 α (20)

0 = 400 + 40 α

400 = – 40 α

α = – 400 / 40

α = – 10 rad/s^{2}

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/s^{2}. 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 (ω_{o}) = 0 rad/s

Angular acceleration (α) = 0.3 rad s^{-2}

The initial angular position (θ_{o}) = 0

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

__Solution :__

The final angular speed after 10 seconds :

ω_{t} = ω_{o} + 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

7. An object rotates with the initial speed of 4 rad/s and the angular acceleration is 0.5 rad/s^{2}. What is the speed of object after 4 seconds.

__Known :__

The initial angular speed (ω_{o}) = 4 rad/s

Angular acceleration (α) = 0.5 rad/s^{2}

Time interval (t) = 4 seconds

__Wanted :__ Object’s speed after 4 seconds (ω_{t})

__Solution :__

ω_{t} = ω_{o} + α t

ω_{t} = 4 + (0.5)(4)

ω_{t} = 4 + 2

ω_{t} = 6 rad/s

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

The correct answer is B.

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.

The correct answer is C.

**Ebook PDF nonuniform circular motions sample problems with solutions 47.59 KB**

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