Assignment 2, solutions

Problem 1:

Sandra, who has a mass m = 40 kg stands on a M = 28 kg flatboat.  Her distance to the shore is 9.4 m.  She walks 2.6 m along the boat toward the shore and then stops.  How far is she away from the shore?  Assume there is no friction between the boat and the water.

Solution:

Concepts, principles, relations that apply to the problem:
Newton’s third law
Why do they apply?
As she walks a distance d1 towards the shore the boat moves a distance d2 away from the shore.
How do they apply?
With respect to the shore:  md1 = Md2.  d1 = (2.6 - d2).
40 kg(2.6 – d2) = 28 kg d2,
68 kg d2 = 40 kg 2.6m,
d2 = 1.53 m.
She has walked 1.07 m towards the shore.  Therefore she is now 8.33 m from the shore.
Details of the calculation:
None

Problem 2:

A small mass slides without friction down the loop track shown in the figure below.
(a)  Show that the speed at point B must be at least as large as (gR)1/2 if the mass does not fall away from the track.
(b) What must be the height h required to achieve the speed found in (a)?  Give your answers in terms of R.

 

Solution:

Concepts, principles, relations that apply to the problem:
Energy conservation, centripetal acceleration
Why do they apply?
We need to make sure that the normal force does not vanish at any position on the upper half of the track except possibly on the top, at q = 90o.  The normal force must be partly responsible for the centripetal acceleration v2/R, otherwise we have projectile motion.
How do they apply?
(a)  On the upper half of the track, the centripetal acceleration needed to keep the mass moving on the circle is partly provided by gravity and partly by the normal force.  When the normal force vanishes, then the mass behaves like a free particle in a gravitational field.  We need to make sure that the normal force does not vanish at any position on the upper half of the track except possibly on the top, at q = 90o.
mv2/R = mgsinq + N.  We need N > 0, therefore we need v2/R > gsinq for all q not equal to 90o.  We need v ³ (gR)1/2.
(b)  Energy conservation:  mgh = mg2R + ½mv2, with v2 = gR.
h = 2R + ½ R = 2.5 R.
Details of the calculation:
 None

Problem: 3

A uniform carpenter's square has the shape of an L, as shown in the figure.  Locate the center of mass relative to the origin of the coordinate system.

Solution:

Concepts, principles, relations that apply to the problem:
The center of mass
Why do they apply?
We are asked to find the CM of an object.
How do they apply?
We can think of the system as being made up of two subsystems, as shown in the figure.

The CM of the left subsystem lies at xCM = 2 cm, yCM = 9 cm.  The CM of the right subsystem lies at xCM = 8 cm, yCM = 2 cm.  If the mass of a 1 cm by 1 cm square is 1 unit, then the mass of the left subsystem is 72 units and the mass of the right subsystem is 32 units.  We find the CM of the system by treating each subsystem as a separate particle, with all its mass concentrated at its center of mass.  Then we have for the whole system

xCM = (72 units 2 cm + 32 units 8 cm)/(104 units) = 3.85 cm,

yCM = (72 units 9 cm + 32 units 2 cm)/(104units) = 6.85 cm.

The CM of the system lies outside of the system.  For irregular-shaped objects it is quite common for the CM to lie outside the system.  This special point outside the system responds to external forces as if the total mass of the system were concentrated there.

Details of the calculation:
None

Problem 4:

A solid iron cylinder (density = 7.87 g/cm3) of radius r = 5 cm and length l = 20 cm rolls down a ramp which has an incline of 20o (no sliding).  The initial height is 3m above ground.
(a)  What is the magnitude of the linear acceleration at half the height?
(b)  The cylinder arrives at the bottom of the ramp. What is the angular momentum of the cylinder about its central axis if it suddenly lifted up from the ground at the two ends of this axis?

Solution:

Concepts, principles, relations that apply to the problem:
Newton's second law, energy conservation, rolling
Why do they apply?
We use Newton's 2nd law to find the linear acceleration.  The rolling constraint is used to eliminate the frictional force in the equation.
How do they apply?
(a) mgsin20o – f = ma, rf = Ia,  a = a/r (rolling).
mgsin20o = (m + I/r2)a.  a = mgsin20o/(m + I/r2) at any position on the ramp.
I = ½ mr2, m = rpr2l = 7.87 g/cm3* p*25 cm2*20 cm = 12.36 kg.
a = 9.8m/s2 * sin20o/1.5 = 2.23 m/s2.
Details of the calculation:
(b)  Rolling:  w = v/R.  Energy conservation: ½ mv2 + ½ Iw2 = mgh,  (3/4)mv2 = mgh.
v = (4gh/3) = 4*9.8 m/s = 39.2 m/s.
L = mrv at the bottom of the ramp.  Lifting the cylinder at the two ends does not exert a torque about the axis, therefore L stays constant.
L = 12.36 kg*0.05m* 39.2 m/s = 24.2 kgm2/s

Problem 5:

A homogeneous thin rod of mass m and length 2a slides on a smooth, horizontal table, one end being constrained to slide without friction in a fixed straight line.  It is initially at rest, with its extension normal to the line, when it is struck at the free end with an impulse Q parallel to the line.
(a)  Determine the initial motion of the rod.
(b)  Show that the force exerted by the line on the rod is given by 
Q2sinq/[ma((4/3) -sin2q)2], where q is the angle between the rod and the line.

Solution:

Concepts, principles, relations that apply to the problem:
Newton's second law:  F = dP/dt, t = dL/dt.  The linear and the angular motion are coupled by a constraint.
Why do they apply?
The force delivering the impulse acts at a point away from the CM.  This results in a torque about the CM.  The force of constraint also acts away from the CM and therefore can produce a torque.
How do they apply?

The impulse is in the x-direction.  It will result in momentum P of the CM and angular momentum L about the CM.  The force of constraint is in the y-direction.  It can change the y-component of P and it can change L.
The y coordinate of the CM and the angular coordinate q are linked by constraints.
Details of the calculation:
(a) initial motion of CM: 
P = Qi, V = (Q/m)i, initial motion about CM:  L = -Qak, w = -(Qa/I)k.
moment of inertia of the rod about the CM:    .
(b) subsequent motion:  There is no force in the x-direction, Vx = Q/m = constant.
There is a force in the y-direction which produces linear acceleration of the CM and angular acceleration about the CM. 
d2y/dt2 = -F/m, dL/dt = F cosq ak = I(d2q/dt2)k.
Therefore d2y/dt2 = -I(d2q/dt2)/(m cosq a).  (y is the y-coordinate of the CM.)
But y and q are linked by constraints. 
y = a sinq,  d2y/dt2 =  - a sinq (dq/dt)2 + a cosq (d2q/dt2).
Therefore - a sinq (dq/dt)2 + a cosq (d2q/dt2) = -I(d2q/dt2)/(m cosq a),
d2q/dt2 = cosq sinq (dq/dt)2/[(4/3) - sin2q]

We are supposed to show that F = Q2sinq/[ma((4/3) - sin2q)2].
Using F cosq ak = I(d2q/dt2)k, we are supposed to show that  d2q/dt2 = (Q2/I)cosqsinq/[ma((4/3) -sin2q)2].
[To solve this differential equation guess (dq/dt) = A/((4/3) - sin2q)1/2 and verify the solution.  This is an obvious guess, since the solution for F is given.  A is determined from the initial conditions.
At q = 90o, dq/dt = Qa/I = 3Q/ma = 31/2A.  A = 31/2Q/ma.]
F = I(d2q/dt2)/[cosq a] = [ma/(3cosq)][A2sinq cosq/((4/3) -sin2q)2] = Q2sinq/[ma((4/3) -sin2q)2]

Problem 6:

Two identical uniform cylinders of radius R each are placed on top of each other next to a wall as shown.  After a disturbance, the bottom cylinder slightly moves to the right and the system comes into motion.  Find the maximum subsequent speed of the bottom cylinder.  Neglect friction between all surfaces.

Solution:

Concepts, principles, relations that apply to the problem:
Energy conservation
Why do they apply?
No friction  à  no torque
The cylinders will not rotate.
Assume a constraint.  The cylinders will stay in contact.
Calculate the velocity of the bottom cylinder as a function of X, the horizontal position of its center, using energy conservation.
Calculate its acceleration a function of X.  At the horizontal position X for which dvx/dt = 0, the acceleration changes sign.  The force of constraint now must pull on the cylinders together.  This constraint cannot be met.  The cylinders will not stay in contact.  The lower cylinder will slide to the right with the horizontal velocity it has at the horizontal position X for which dvx/dt = 0.  Because its acceleration and its velocity have both pointed in the positive x-direction before the cylinder reached that position X, the speed the lower cylinder has at that position is its maximum speed.
How do they apply?
 

Let M be the mass of each cylinder and R the radius.  Assume the cylinders stay in contact.
Energy conservation:
Mg(2R – 2Rsinq) = ½ M(vx2 + vy2)
The upper cylinder is moving down and the lower cylinder is moving towards the right.
vx2 + vy2 = 4Rg(1 – sinq)
X = 2Rcosq, Y = 2Rsinq,
vx = dX/dt, vy = dY/dt, vx2 + vy2 = 4R2(dq/dt)2
(dq/dt)2 = (g/R)(1 – sinq)
Energy conservation yields dq/dt in terms of the angle q.  Since vx is proportional to dq/dt,  we can now find vx and dvx/dt in terms of q, and therefore find the angle q for which dvx/dt = 0.
vx = -2R sinq dq/dt, vx2 = 4Rg(sin2q - sin3q)
dvx2/dt = (dvx2/dq)(dq/dt) = 2vx(dvx/dt)
dvx2/dt = 4Rg(2sinq – 3sin2q) (dq/dt)
dvx/dt = dvx2/dt = 0 if  2sinq – 3sin2q = 0.
sinq = 2/3, dq/dt = -(g/3R)1/2, vx = (4/3)(gR/3)1/2.
The maximum speed of the bottom cylinder is (4/3)(gR/3)1/2.

Details of the calculation:
None

Problem 7:

A pencil of length L is held vertically with its point on a desk and then allowed to fall over.  Assuming that the point does not slip, find the speed of the eraser just as it strikes the desk.  Compare this with the speed that would result from free fall from a height equal to the length of the pencil.

Solution:

Concepts, principles, relations that apply to the problem:
Energy conservation
Why do they apply?
Gravity is a conservative force,
How do they apply?
We have pure rotation about a point.
(1/2)Iw2 = mgL/2.  I = mL2/3.
w2 = 3g/L.  vtip = wL = (3gL)1/2.
Details of the calculation:
Speed of eraser that falls freely from a height L: v = (2gL)1/2.
[(1/2)mv2 = mgL]