Question

1. A wheel that initially spins at 2 revolutions/sec is braked uniformly to a stop in ½ second. How many revolutions does the wheel make while coming to a stop?

2. If a person has a speed of 3 m/s and is located 2-m from the axis of rotation of a merry-go-round, what is the person's angular velocity?

3. A hockey player with a mass of 46.0 kg is traveling due east with a speed of 2.95 m/s. A second hockey player with a mass of 66.0 kg is moving due south with a speed of 6.95 m/s. They collide and hold on to each other after the collision. Find the direction they travel after the collision given as an angle measured south of east.

4. The Seattle ferris wheel has a radius of 11-m and travels with an angular velocity of 1.5 rad/sec. What is the normal force (apparent weight) acting on a 57-kg person at the bottom of the swing?

5. A 65 kg snowboarder travels down a 37° incline at 6.2 m/s. In order to stop before hitting a tree, the snowboarder quickly turns their board to increase the coefficient of kinetic friction to 1.9. What is the normal force between the snowboarder and the ground?

6. A 65 kg snowboarder travels down a 37° incline at 6.2 m/s. In order to stop before hitting a tree, the snowboarder quickly turns their board to increase the coefficient of kinetic friction to 1.9. What is the friction force between the snowboarder and the ground?

7. A 65 kg snowboarder travels down a 37° incline at 6.2 m/s. In order to stop before hitting a tree, the snowboarder quickly turns their board to increase the coefficient of kinetic friction to 1.9. How far does the snowboarder slide before stopping? Assume the path the snowboarder takes is straight down the slope.

Answer 1

1.

Using 1st rotational kinematic equation:

wf = wi + alpha*t

wi = 2 rev/sec = 2*2*pi rad/sec

wf = 0 rad/sec

t = 0.5 sec, So

alpha = angular acceleration = (wf - wi)/t

alpha = (0 - 4*pi)/0.5 = -8*pi rad/sec^2

Now Using 3rd kinematic equation:

wf^2 = wi^2 + 2*alpha*theta

theta = (wf^2 - wi^2)/(2*alpha) = (0^2 - (4*pi)^2)/(2*(-8*pi)) = pi rad

Now since 1 rev = 2*pi rad, So

theta = pi rad = pi/(2*pi) = 1/2 rev = revolution made by wheel before stopping

2.

Angular velocity = w = V/R

w = (3 m/sec)/(2 m) = 1.5 rad/sec

3.

Since there is no external force applied, So using momentum conservation:

Pi = Pf

Since they grab onto each other after collision, So their final velocity will be same

m1*v1 + m2*v2 = (m1 + m2)*V

m1 = 46 kg and m2 = 66 kg

v1 = initial velocity of m1 = 2.95 m/sec towards east = (+2.95 i) m/sec

v2 = initial velocity of m2 = 6.95 m/sec towards south = (-6.95 j) m/sec

So,

46*(2.95 i) + 66*(-6.95 j) = (46 + 66)*V

V = [135.7 i - 458.7 j]/(46 + 66)

V = 1.21 i - 4.096 j

Now final speed will be:

|V| = sqrt (1.21^2 + (-4.096)^2) = 4.27 m/sec

Direction = arctan (4.096/1.21) = 73.5 deg South of east