Elements of Physics, 131:01 Name________________

November 10, 2003

Test 2.

The working for all solutions must be shown. No credit will be given for an answer with no working. Take g to be 9.8m/s².

1. A luge and its rider, with a total mass of 80 kg, emerges from a downhill track onto a horizontal straight track with an initial speed of 40 m/s. If they stop at a constant deceleration of 2.0 m/s², (a) what magnitude F is required for the decelerating force, (b) what distance d do they travel wile decelerating, and (c) what work W is done on them by the decelerating force? What are (d) F, (e) d, and (f) W for a deceleration of 4.0 m/s²?

(a) The force is given by F = ma. The magnitude of the force is

F = 80 kg × 2.0 m/s² = 160 N

(b) You can get d from the kinematic equations, for example

v² = v₀² + 2a(x-x₀), d = (v² - v₀²)/2a = (-1600 m²/s²)/(-2.0 m/s²) = 400 m

(c) W = Fd = 160N × (-400m) = -64000 J

(the minus sign is there because the distance moved in the direction of the force is negative, but I didn't take off marks if you didn't use it.)

(d) If a is doubled, the force is 320 N

(e) If a is doubled, the distance is 200 m

(f) The work is Fd and is unchanged. The initial kinetic energy of the system was not changed, and this determines the amount of work.

2. A 30 g bullet, with a horizontal velocity of 550 m/s, comes to a stop 15 cm within a solid wall. (a) What is the change in its mechanical energy? (b) What is the magnitude of the average force from the wall stopping it?

(a) The change in the mechanical energy is entirely the change in the kinetic energy

= K_f - K_i = 0 - ½(0.030 kg)(550 m/s)² = -4537.5 J

(b) This equals the work done on the bullet, and the work is done while the bullet moves 15 cm = 0.15 m, so the average force is

4537.5 J / 0.15 m = 30250 N (the question asked for the magnitude, so ignore the minus sign)

3. What are (a) the x coordinate and (b) the y coordinate of the center of mass of the three particle system shown? (c) What happens to the center of mass as the mass of the topmost particle is gradually increased? See Fig 9-22

The formula for each component of the position of the center of mass is of the form

For the x-component

x_cm = (3.0kg × 0 + 6.0kg × 1m + 4.0kg×2m)/(3.0kg + 6.0kg + 4.0kg)

=1.077m

(notice that you have to keep the 3.0kg in the denominator even though it is multiplied by zero in the numerator)

y_cm = (3.0kg × 0 + 6.0kg × 2m + 4.0kg×1m)/(3.0kg + 6.0kg + 4.0kg)

=1.231m

(c) The point of this is that the heaviest mass tends to dominate the sum. If the 6.0kg mass is steadily increased, the center of mass will move towards its coordinates, x = 1m, y = 2m.

4. A 5.0 kg box sled is coasting across frictionless ice at a speed of 8.0 m/s when a 15.0 kg package is dropped into it from above. What is the new speed of the sled?

The horizontal momentum of the system is conserved. The package is dropped vertically so it has zero initial horizontal component of momentum.

(5.0 kg)(8.0 m/s) + (15 kg)(0) = (5.0 kg + 15 kg)v_f

40 kg-m/s = (20 kg)v_f

v_f = (40 kg-m/s)/(20 kg) = 2.0 m/s

5. A wheel has a constant angular acceleration of 4.0 rad/s². During a certain 3.0 s interval, it turns through an angle of 120 rad. Assuming that the wheel starts from rest, how long has it been in motion at the start of this 3.0 s interval?

The angular speed at the beginning of the 3.0 s interval is found from

q - q₀ = w₀t + ½at²

120 rad = w₀(3.0 s) + ½(4.0 rad/s²)(3.0 s)²

The solution of this is w₀ = 34 rad/s

To find how long the wheel was in motion before the 3.0 s interval began, use this as the final angular velocity in

w = w₀ + a t

34 = 0 + (4.0)t

t = 8.5 s

6. A girl of mass M stands on the rim of a frictionless merry-go-round of radius R and rotational inertia I that is not moving. She throws a rock of mass m horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock relative to the ground is v. Afterward, what are (a) the angular speed of the merry-go-round, and (b) the linear speed of the girl?

The angular momentum of the whole system is conserved, and its initial value is zero. After the girl throws the rock, it has angular momentum about the center of the merry-go-round of mvR. The merry-go-round has angular momentum Iw and the girl has angular momentum MR²w. (Her rotational inertia is MR² and she has the same angular speed as the merry-go-round.) Therefore

Iw + MR²w + mvR = 0

(I + MR²)w = -mvR

(a) w = -mvR/(I + MR²)

(The minus sign is not important. I could have said that the angular momentum of the rock was -mvR.)

(b) The linear speed of the girl is

V = wR = mvR²/(I + MR²)

7. A uniform sphere of mass m and radius r is held in place by a massless rope attached to a frictionless wall a distance L above the center of the sphere. Find (a) the tension in the rope and (b) the force on the sphere by the wall.

The free-body diagram for the sphere is

For the vertical forces to be in equilibrium

W = mg = T cos(q)

so T = mg/cos(q)

and for the horizontal forces to be in equilibrium

N = T sin(q) = mg tan(q)

From the left hand picture, tan(q) = r/L and cos(q) = L/(L²+r²)^1/2

so

T = (mg/L)(L²+r²)^1/2

N = mgr/L

8. Two particles, each with mass m, are fastened to each other and to a rotation axis at O, by thin rods, each with length d and mass M. The combination rotates around the rotation axis with angular velocity . In terms of these symbols, and measured about O, what are the combination's (a) rotational inertia and (b) kinetic energy? The rotational inertia of a rod about one end is (1/3)ML².

(a) the length of the rod is 2d, and its total mass is 2M, so its rotational inertia is (2M)(2d)²/3 = 8Md²/3

The particle at the middle of the rod has a rotational inertia of md², and the particle at the end of the rod has a rotational inertia of m(2d)² = 4md².

The total rotational inertia is therefore

I = 8Md²/3 + 5md²

(b) The kinetic energy is

K = ½Iw² = ½(8Md²/3 + 5md²)w²