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Homework 6: Angular Velocity and Acceleration Calculations, Study notes of Calculus

Solutions to various problems related to angular velocity and acceleration. Topics include calculating angular velocity from revolutions per minute, converting radians to degrees, finding centripetal and tangential acceleration, and determining if an object will slide off a rotating surface. Each problem involves real-world scenarios such as bicycle wheels, crankshafts, and astronaut centrifuges.

What you will learn

  • What is the tangential acceleration of an object moving in a circle with a given angular acceleration and radius?
  • At what angular velocity will a coin slide off a rotating turntable with a given radius and coefficient of friction?
  • What is the angular velocity of a wheel spinning at 56 revolutions per minute?
  • How do you convert radians to degrees?
  • What is the centripetal acceleration of an object moving in a circle with a given radius and angular velocity?

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HW 6 1
Homework 6
1. 1 revolution = 2πradians and 1 minute = 60 seconds. Therefore, 56 revolutions per minute = 56 ×2π/60
rad/s= 5.86 rad/s.
2. a)x=, or θ=x/r radians. Therefore, θ= 5000/4000 = 1.25 radians.
1 radian=180, therefore, 1.25 radians = 1.25 ×180/2π= 71.62.
b) Angular velocity = angular distance/time. The plane takes 8 hours, thus, ω= 1.25/(8 ×60 ×60) =
4.34 ×105rad/s.
3. Your roommate is working on his bicycle and has the bike upside down. He spins the 56 cm (or 0.56 m)
diameter wheel, and a pebble in the tread goes by 6 times every second. Thus, the wheel’s angular speed
is 6 rev/s or 6 ×2π= 37.7 rad/s. Tangential speed, v=, thus, v= (0.56/2) ×37.7 = 10.56 m/s.
radial acceleration,
ac=v2
r=10.562
0.28 = 397.942 m/s2
4. Mass of the ball = 28 g = 0.028 kg
The track is 42 cm in diameter or 42/2 = 21 cm in radius.
Speed of the ball = 60 rpm = 60 ×2π/60 rad/s = 6.283 rad/s.
Centripetal acceleration of the ball= ac=ω2r= 6.2832×0.21 = 8.29 m/s2.
The force for this acceleration comes from the track. F=mac= 0.028 ×8.29N= 0.232N.
5. Diameter of the crankshaft = 3.0 cm. Therefore, radius = 3.0/2 = 1.5 cm = 0.015 m.
Initial rotation speed = 2400 rpm = 2400 ×2π/60 rad/s = 251.327 rad/s.
The wheel comes to a stop in 1.5 s.
a) Therefore, its angular acceleration can be calculated using,
ωf=ωi+αt
0 = 251.327 + α×1.5
α=167.552 rad/s2
Now, tangential acceleration, at=αr =167.552 ×0.015 = 2.513 m/s2.
b) The angular distance it covers before coming to a stop is given by,
θ=ωi+ωf
2t
θ=251.327 + 0
21.5 = 188.496 rad
or θ= 188.496/(2π) = 30 rev
6. Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 34 s to
speed up from rest to its top speed of 1 rotation every 1.3 s. The radius of the circle in which the astronaut
moves is 6.8 m.
a) The astronaut’s top angular speed is 1 revolution per 1.3 s. Thus, its speed is 1 ×2π/1.3 rad/s =
pf3

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Homework 6

  1. 1 revolution = 2π radians and 1 minute = 60 seconds. Therefore, 56 revolutions per minute = 56 × 2 π/ 60 rad/s= 5.86 rad/s.
  2. a)x = rθ, or θ = x/r radians. Therefore, θ = 5000/4000 = 1.25 radians.

1 radian=180/π, therefore, 1.25 radians = 1. 25 × 180 / 2 π = 71. 62 ◦. b) Angular velocity = angular distance/time. The plane takes 8 hours, thus, ω = 1. 25 /(8 × 60 × 60) =

  1. 34 × 10 −^5 rad/s.

  2. Your roommate is working on his bicycle and has the bike upside down. He spins the 56 cm (or 0.56 m) diameter wheel, and a pebble in the tread goes by 6 times every second. Thus, the wheel’s angular speed is 6 rev/s or 6 × 2 π = 37.7 rad/s. Tangential speed, v = rω, thus, v = (0. 56 /2) × 37 .7 = 10.56 m/s. radial acceleration, ac = v^2 r

= 397.942 m/s^2

  1. Mass of the ball = 28 g = 0.028 kg The track is 42 cm in diameter or 42/2 = 21 cm in radius. Speed of the ball = 60 rpm = 60 × 2 π/60 rad/s = 6.283 rad/s. Centripetal acceleration of the ball= ac = ω^2 r = 6. 2832 × 0 .21 = 8.29 m/s^2. The force for this acceleration comes from the track. F = mac = 0. 028 × 8. 29 N = 0. 232 N.
  2. Diameter of the crankshaft = 3.0 cm. Therefore, radius = 3.0/2 = 1.5 cm = 0.015 m. Initial rotation speed = 2400 rpm = 2400 × 2 π/60 rad/s = 251.327 rad/s. The wheel comes to a stop in 1.5 s. a) Therefore, its angular acceleration can be calculated using,

ωf = ωi + αt

0 = 251.327 + α × 1. 5 α = − 167 .552 rad/s^2 Now, tangential acceleration, at = αr = − 167. 552 × 0 .015 = − 2 .513 m/s^2. b) The angular distance it covers before coming to a stop is given by,

θ = ωi + ωf 2

t

θ =

1 .5 = 188.496 rad

or θ = 188. 496 /(2π) = 30 rev

  1. Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 34 s to speed up from rest to its top speed of 1 rotation every 1.3 s. The radius of the circle in which the astronaut moves is 6.8 m. a) The astronaut’s top angular speed is 1 revolution per 1.3 s. Thus, its speed is 1 × 2 π/ 1 .3 rad/s =

4 .833 rad/s. We can calculate his angular acceleration using, ωf = ωi + αt, or 4.833 = 0 + α × 34, or α = 0.142 rad/s^2. The tangential acceleration, at = αr = 0. 142 × 6 .8 = 0.9667 m/s^2. b) When the device is rotating at its top speed, the centripetal acceleration of the astronaut is given by ar = ω^2 r = 4. 8332 × 6 .8 = 158.848 m/s^2. In terms of g, it is, 158. 848 / 9 .8 = 16. 21 g.

  1. A 485 g (0.485 kg) ball swings in a vertical circle at the end of a 2.1 m-long string. At the bottom of the circle, the tension in the string is 21 N. The weight of the ball is mg = 0. 485 × 9 .8 = 4. 753 N. Therefore, the net force on the ball is T − mg = 21 − 4 .753 = 16. 247 N. The acceleration produced by this force is the centripetal acceleration of the ball.

ac = F/m = 16. 247 / 0 .485 = 33.5 m/s^2

Now, centripetal acceleration, ac =

v^2 r 33 .5 = v^22. 1 v = 8.387 m/s.

  1. A 5.0 g is placed 15 cm (0.15 m) from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of μs = 0.8 and μk = 0.5. The turntable very slowly speeds up to 60 rpm. Does the coin slide off? That the turntable speeds up very slowly means that we can ignore any tangential acceleration of the coin. All we have to consider is the centripetal acceleration. The force for centripetal acceleration is provided by the force of static friction. mω^2 r = μsN ⇒ mω^2 r = μsmg ⇒ ω^2 r = μsg ⇒ ω^2 = 0. 8 × 9. 8 / 0 .15 = 52. 267 or ω = 7.23 rad/s. Converting to rpm, 7.23 rad/s=7. 23 / 2 π × 60 rpm = 69 rpm. Thus, at 60 rpm the coin will not slide off the turntable.
  2. A car drives over the top of a hill that has a radius of 38 m. At the speed at which the car loses contact, the normal force from the ground vanishes. Thus, the acceleration due to gravity is the centripetal acceleration of the car, or v^2 r

= g

v^2 = 38 × 9 .8 = 372. 4 v = 19.298 m/s.

  1. A 450 g (0.45 kg) steel block rotates on a steel table while attached to a 1.2 m long hollow tube. A thrust force of 4.8 N perpendicular to the tube is exerted by compressed air. The maximum tension the tube can withstand without breaking is 50 N. The tension in the tube provides the centripetal acceleration of the block. Thus, the angular velocity of the block at the instant the tube breaks off is given by,

ac =

Tmax m

= ω^2 r

50

  1. 45 = ω^2 × 1. 2

ω^2 = 92. 593