Rotational Kinematics Flashcards
A child sitting 1.14 m from the center of a merry-go-round moves with a speed of 1.40 m/s.
(a) Calculate the centripetal acceleration of the child.
(b) Calculate the net horizontal force exerted on the child (mass = 25.0 kg).
(a) 1.72 m/s^2 toward the center (b) 42.98 N toward the center
Centripetal accel = v^2/r = (1.05^2)/1.08
Centripetal Force = F= mv^2/r= (25)(1.05^2)/1.08
Suppose the space shuttle is in orbit 320 km from the Earth’s surface, and circles the Earth about once every 90.9 minutes. Find the centripetal acceleration of the space shuttle in its orbit. Express your answer in terms of g, the gravitational acceleration at the Earth’s surface.
0.91 g
centrip accel = v^2/r
we need to find the speed and the radius
the speed is the distance traveled/time
the distance is the radius of the orbit, or 2 pi r; the time is the period P
so v=2 pi r/P and v^2=4 pi^2 r^2/P^2
then a=v^2/r = 4 pi^2 r/P^2
r=radius of earth + 320 km
= 6400 km + 320 km = 6720 km
= 6.872x10^6m
P=90.9 mins = 90.9 minsx60 sec/min = 5454s
a=4 pi^2 (6.72x10^6)/(5454)^2
= 8.92 m/s/s => 8.92/9.8 = 0.91 g
What is the magnitude of the acceleration of a speck of clay on the edge of a potter’s wheel turning at 48 rpm (revolutions per minute) if the wheel’s diameter is 31 cm?
3.92 m/s^2
R = radius of wheel = 31/2 = 15.5 cm = 0.155 m
angular velocity = w = 48 rpm = 48/60 rps = 2π(48/60) rad/s = 5.03 rad/s
linear velocity of speck = V = wR = (5.03)(0.155) = 0.780 m/s
Centripetal acceleration = V²/R = (0.780)²/0.155 = 3.92 m/s² ANS
A ball on the end of a string is revolved at a uniform rate in a vertical circle of radius 80.0 cm, as shown in Fig. 5-33. Its speed is 3.90 m/s and its mass is 0.300 kg
(a) Calculate the tension in the string when the ball is at the top of its path.
(b) Calculate the tension in the string when the ball is at the bottom of its path.
(a) 2.764 N
(b) 8.644 N
Fc = mv² / r Fc = 0.300(4.10²) / 0.85 Fc = 5.93294118 N
a) Ft = Fc - Fg
Ft = 5.93294118 - 0.300(9.81)
Ft = 2.98994118 N
Ft = 2.99 N
b) Ft = Fc + Fg
Ft = 5.93294118 + 0.300(9.81)
Ft = 8.87594118 N
Ft = 8.88 N
The tension is greatest at the bottom of the path.
A 0.40 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.1 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 56 N, what is the maximum speed the ball can have?
12.41 m/s
Ft (tension force) = Fc (centripetal force)
Ft = 56N m = 0.4kg r = 1.1m v = ?
Ft = mv^2 / r v^2 = Ftr / m v = sqrt (Ftr / m) v = 12.41 m/s
A car at the Indianapolis-500 accelerates uniformly from the pit area, going from rest to 335 km/h in a semicircular arc with a radius of 200 m.
(a) Determine the tangential acceleration of the car when it is halfway through the turn, assuming constant tangential acceleration.
(b) Determine the radial acceleration of the car at this time.
(c) If the curve were flat, what would the coefficient of static friction have to be between the tires and the roadbed to provide this acceleration with no slipping or skidding?
(a) m/s2
(b) m/s2
(c)
What are the following angles expressed in radians? (Give as numerical values and as fractions of π.)
(a) 30° = __π = __rad
(b) 58° = __π = __rad
(c) 90° = __π = __rad
(d) 360° = __π = __rad
(e) 470° = __π = __rad
(a)
(b)
(c)
(d)
(e)
The blades in a blender rotate at a rate of 7300 rpm. When the motor is turned off during operation, the blades slow to rest in 4.0 s. What is the angular acceleration as the blades slow down?
rad/s2
A child rolls a ball on a level floor 3.5 m to another child. If the ball makes 11.0 revolutions, what is its diameter?
m
A grinding wheel 0.26 m in diameter rotates at 2500 rpm.
(a) Calculate its angular velocity in rad/s.
(b) What are the linear speed and acceleration of a point on the edge of the grinding wheel?
linear speed =
radial acceleration =
(a) m/s
(b) linear speed = m/s
radial acceleration = m/s2
A rotating merry go-round makes one complete revolution in 3.8 s (Figure 8-38).
Figure 8-38
(a) What is the linear speed of a child seated 0.7 m from the center?
(b) What is her acceleration (give components)?
tangential acceleration =
radial acceleration = (towards or away from center?)
(a)
(b) tangential acceleration = m/s2
radial acceleration = m/s2
How fast (in rpm) must a centrifuge rotate if a particle 12.0 cm from the axis of rotation is to experience an acceleration of 110,000 g’s?
rpm
A centrifuge accelerates from rest to 16,000 rpm in 230 s. Through how many revolutions did it turn in this time?
revolutions
An automobile engine slows down from 3800 rpm to 1200 rpm in 4.5 s.
(a) Calculate its angular acceleration, assumed constant.
(b) Calculate the total number of revolutions the engine makes in this time.
(a) rad/s2
(b) rev
A wheel 40.0 cm in diameter accelerates uniformly from 215 rpm to 350 rpm in 6.5 s. How far will a point on the edge of the wheel have traveled in this time?
m
