November 14, 2007
PHY2053
Quiz 8 (Chapter 8)
Name:
UFID:
1. (5 pts) A 20.0m, 40.0kg uniform ladder rests against a frictionless wall, making an
angle of 60º with the horizontal. Find the minimum coefficient of static friction between
ladder and ground so that a 80.0 kg firefighter can reach the top of the ladder.
Since each component of the net force equals zero, we get
X: n-mg-Mg = 0 ⇒ n = (m+M)g = (80+40)9.8 = 1176 N, Y: N-f = 0 ⇒ f = N,
where m = 80kg, M = 40kg, n/N are the normal force from the ground/the wall.
Now we calculate the torque around the bottom of the ladder when the firefighter
reaches the top:
mgLcosθ+Mg(L/2)cosθ-NLsinθ = 0
⇒ N = (gcosθ(m+M/2))/sinθ = 9.8cos60º(80+40/2)/sin60º = 565.8 N
If the coefficient is the minimum, the friction reaches the maximum static friction when
the firefighter reaches the top, thus
μ = f/n = N/n = 565.8/1176 = 0.481
2. (5 pts) (5 pts) A 4.00 kg bucket is attached to a light cord which is wrapped around a
6.00 kg cylindrical reel. The radius of the reel is 0.500 m and the moment of inertia of
the reel is given by I = (1/2)MR². The bucket falls from rest into a well. What is the
tension in the string?
We take the positive direction of the bucket downward and that of the reel
counterclockwise. Then αR = a. The equations of motion for the bucket and the reel are
given by:
ma = mg-T
Iα = TR ⇒ (1/2)MR²(a/R) = TR ⇒ (1/2)Ma = T
Adding these two equations, we get
(m+(1/2)M)a = mg ⇒ a = mg/(m+0.5M) = 4×9.8/(4+0.5×6) = 5.60 m/s²
We use the second equation to calculate the tension:
T = (1/2)Ma = 0.5×6×5.6 = 16.8 N
3. (5 pts) A cheerleader has a baton made of a 0.300 kg, 0.500 m solid rod and two 0.100
kg masses attached both ends of the rod. Calculate the moment of inertia of the baton
around the axis through the center. The moment of inertia of a solid rod around the
center is given by I = (1/12)ML².
The moment of inertia of the baton is the sum of the moment of inertia of the rod and
that of the masses, thus
I(baton) = (1/12)ML² + 2m(L/2)² = 1/12×0.3×0.5²+2×0.1×0.25² = 1.88×10ˉ² kg m²
4. (5 pts) A 0.200 kg ball with a radius 0.100 m rolls on a horizontal surface. Then the
ball hits a curb with a height of 0.100 m. Find the minimum angular velocity of the ball
before it hits the curb so that it rolls up the curb without slipping. The moment of
inertia of the ball around the axis through the center is (2/5)MR² and that around the
axis through the surface is (7/5)MR². (Hint: The angular momentum of the ball is
conserved during the collision.)
Since the angular momentum is conserved during the collision, we get
Iω = I’ω’ ⇒ (2/5)MR²ω = (7/5)MR²ω’ ⇒ ω = (7/2)ω’
After the collision, the ball rotates around the point of contact and the energy is
conserved. Besides, if the initial angular velocity is the minimum, the final angular
velocity is zero. Thus we get
(1/2)(7/5)MR²ω’² = MgR ⇒ ω’ = √(10g/(7R))
Substituting this into the first equation, we get
ω = (7/2) √(10g/(7R)) = 3.5×√(10×9.8/(7×0.1) = 41.4 rad/s