__Work__

A cyclist intends to cycle up a 8^{o}
hill whose vertical height is 150 m with constant speed. If each complete revolution of the pedals
moves the bike 6 m along its path, calculate the average force that must be
exerted on the pedals tangent to their circular path. Neglect work done by
friction and other losses. The pedals turn in a circle of diameter 30 cm. The
total mass of the cyclist and his bike is 100 kg.

Solution:

- Concepts:

Work and energy - Reasoning:

The work done by the cyclist changes the potential energy of the system. - Details of the calculation:

The net work done by the cyclist is during one revolution of the pedals is

W = mg∆h = 100*9.8*6*sin(8^{o}) J = 818 J.

During one one revolution the pedal travels a distance s.

W = <F>*s = <F>*2π(0.15 m) = (0.942 m)*<F>

<F> = (818 J)/( (0.942 m) = 868 N

__Conservation of energy__

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:

Energy conservation - Reasoning:

Gravity is a conservative force. We have pure rotation about a point. - Details of the calculation:

½Iω^{2}= mgL/2. I = mL^{2}/3.

ω^{2}= 3g/L. v_{tip}= ωL = (3gL)^{½}.

Speed of eraser that falls freely from a height L: v = (2gL)^{½}.

[½mv^{2}= mgL]

Two thin beams of mass m and length l are connected by a frictionless hinge and thread.

The system rests on a smooth surface. At t = 0 the thread is cut. Neglect
the mass of the hinge and the treat.

(a) Find the speed of the hinge when it hits the floor.

(b) Find the time it takes for the hinge to hit the floor in terms of an
integral.

Solution:

- Concepts:

Energy conservation - Reasoning:

The system is frictionless. The symmetry of the problem tells us that the hinge will fall vertically. The potential energy of the two beams will be converted into translational kinetic energy of their CMs and rotational kinetic energy about their CMs. - Details of the calculation:

(a) For a given angle θ, the CMs of beams 1 and 2 are at positions

x_{1}= ½lcosθ, y_{1}= ½lsinθ, x_{2}= -½lcosθ, y_{2}= ½lsinθ.

Their velocity components are

dx_{1}/dt = -½lsinθdθ/dt, dy_{1}/dt = ½lcosθdθ/dt, dx_{2}/dt = ½lsinθdθ/dt, dy_{2}/dt = ½lcosθdθ/dt.

The speed of the CM of each rod is v = [¼l^{2}(dθ/dt)^{2}]^{½}= ½l|dθ/dt|

and its angular speed about its CM is is |dθ/dt| = ω = 2v/l.

The speed of the hinge is v_{h}= 2v.

The moment of inertia if each rod about its CM is I = ml^{2}/12.

The kinetic energy of the system is 2[½mv^{2}+ ½Iω^{2}]

= ¼ml^{2}(dθ/dt)^{2}+ (1/12)ml^{2}(dθ/dt)^{2}= ⅓ml^{2}(dθ/dt)^{2}.

The potential energy of the system is mglsinθ.

Energy conservation: ⅓ml^{2}(dθ/dt)^{2}+ mglsinθ = mglsin(30^{o}) = ½mgl.

(dθ/dt)^{2}= (3g/l)(0.5 - sinθ).

When θ = 0 the |dθ/dt| = (3g/(2l))^{½}, v_{h}= l|dθ/dt| = (3lg/2))^{½}.

This is the speed with which the hinge hits the floor.

(b) dt = dθ/[dθ/dt], t = -∫_{30 deg}^{0}dθ/[(3g/2l)(1 - 2sinθ)]^{½}= (2l/3g)^{½}∫_{0}^{30 deg}dθ/(1 - 2sinθ)^{½}.

Galileo wants to demonstrate that unequal masses fall with a uniform
acceleration by dropping two spherically symmetric masses simultaneously from
the same height z. Ideally we know that they should hit the ground at the same
time, t = (2z/g)^{½}, where g is the acceleration due to gravity.

Unfortunately for Galileo, the surface of the first sphere is slightly
irregular, so the otherwise negligible air drag induces a rotation with an
angular frequency proportional to the velocity, ω = εv_{1}. The second
sphere does not rotate. Assume that ε is constant and ignore possible heating
of the air by friction.

Find the ratio of impact times t_{1}/t_{2} of the two spheres.
The answer should involve only the coefficient ε and the mass m_{1} and
moment of inertia I_{1} of the first sphere.

Solution:

- Concepts:

Energy conservation - Reasoning:

Gravitational potential energy is converted into kinetic energy. The rotating sphere has translational as well as rotational kinetic energy. - Details of the calculation:

For the non-rotating sphere we have as a function of the distance z' below the point of release

mgz' = ½mv^{2}, v^{2}(z') = 2gz'. This yields t(z') = (2z'/g)^{½}.

For the rotating sphere we have as a function of the distance z' below the point of release

m_{1}gz' = ½m_{1}v_{1}^{2}+ ½I_{1}ω^{2}= ½m_{1}v_{1}^{2}+ ½I_{1}ε^{2}v_{1}^{2}= ½(m_{1}+ ε^{2}I_{1})v_{1}^{2},

v_{1}^{2}= 2g[m_{1}/(m_{1}+ ε^{2}I_{1})]z'.

This yields t_{1}(z') = (2z'/g')^{½}, with g' = g[m_{1}/(m_{1}+ ε^{2}I_{1})].

Therefore t_{1}/t_{2}= (g/g')^{½ }= [(m_{1}+ ε^{2}I_{1})/m_{1})]^{½}.

Find the fraction of the kinetic energy that is
translational and rotational when

(a) a hoop

(b) a disc and

(c) a sphere rolls down an inclined plane of height h. Find the velocity at
the bottom in each case. Compare with a block sliding without friction down the
plane.

Solution:

- Concepts:

The moment of inertia, energy conservation, rolling - Reasoning:

Energy conservation: T_{f}= mgh = ½mv^{2}+ ½Iω^{2}

For rolling ω = v/r, so T_{f}= ½v^{2}[m + I/r^{2}] = mgh. - Details of the calculation:

The moments of inertia of the hoop, disk, and sphere with radius r are

I_{h}= mr^{2}, I_{d}= mr^{2}/2 and I_{sp}= 2mr^{2}/5, respectively.

T_{kin}/T_{f}= m/[m + I/r^{2}], T_{rot}/T_{f}= (I/r^{2})/[m + I/r^{2}] = I/[mr^{2}+ I]

v = [2mgh/(m + I/r^{2})]^{½ }For a block T_{rot}= 0, T_{kin}/T_{f}= 1, v = (2gh)^{½ }(a) hoop: T_{kin}/T_{f}= T_{rot}/T_{f}= ½, v = (gh)^{½}= ½^{½}v_{block}.

(b) disk: T_{kin}/T_{f}= ⅔, T_{rot}/T_{f}= ⅓, v = (4gh/3)^{½}= ⅔^{½}v_{block}.

(c) sphere: T_{kin}/T_{f}= 5/7, T_{rot}/T_{f}= 2/7, v = (10gh/7)^{½}= (5/7)^{½}v_{block}.

A sphere of radius R and uniformly distributed mass m is released at rest and
begins moving down an inclined plane which makes an angle q with the horizontal.
The sphere reaches a final position as indicated, with its center at ∆z = -h
relative to its initial position. The gravitational force on the sphere is
**F** = -mg **k**.

(a) If the surface is frictionless, so the sphere slips without
rotating,

(i) what is the time t required to go
from the initial to the final position and

(ii ) what is
the final velocity of the sphere?

(b) Now assume that the surface has sufficient friction so that the sphere rolls without slipping. In that case, what is the final velocity of the central point of the sphere?

Solution**:**

- Concepts:

Newton's 2^{nd}law, energy conservation - Reasoning:

We have a sphere sliding or rolling down an inclined plane. - Details of the calculation:

(a) F = ma = m d^{2}s/dt^{2}= mgsinθ. d^{2}s/dt^{2}= gsinθ. v = ds/dt = gsinθt, s = ½gsinθt^{2}.

The sphere travels a distance s = h/sinθ. h/sinθ = ½gsinθt^{2}. t^{2}= 2h/(gsin^{2}θ).

The time t required to go from the initial to the final position is t = (2h/g)^{½}(1/sinθ).

mgh = ½mv^{2}, v = (2gh)^{½}is the final velocity of the sphere.

(b) ½mv^{2}+ ½Iw^{2}= mgh, I = 2mR^{2}/5 for a uniform sphere, w = v/R for rolling.

½mv^{2}+ mv^{2}/5 = mgh, 7v^{2}/10 = gh, v = (10gh/7)^{½}is the final velocity of the central point of the sphere.

A cylindrical shell rolls without slipping down an incline as shown below. If it starts from rest, how far must it roll along the incline to obtain a speed v?

Solution:

- Concepts:

Energy conservation, rolling - Reasoning:

Gravity is a conservative force.

Rolling: ω = v/r. Energy conservation: ½mv^{2}+ ½Iω^{2}= mgΔh. - Details of the calculation:

I = mr^{2}, therefore v^{2}= gΔh.

Δh = Δs*sinθ, therefore Δs = v^{2}/(g sinθ) is the distance the cylindrical shell must roll.

A cylinder of mass m, radius r, and length l rolls straight down an inclined plane of length L and angle 30° with respect to the horizontal. A sphere of mass M, and radius R rolls down another inclined plane, also of length L but at an angle θ with respect to the horizontal. Each object is solid, of uniform density, and rolls without slipping. Find an expression for the angle θ for which the two objects reach the bottom at the same time after they are released from rest at the same time.

Solution:

- Concepts:

Rigid-body motion, energy conservation, rolling - Reasoning:

The accelerations should be equal for the objects are to reach the bottom at the same time; the lengths L of the inclines does not matter. - Details of the calculation:

The moments of inertia are I_{cyl}= ½mr^{2}, I_{sph}= (2/5)Mr^{2}. The length of the cylinder does not matter.

| potential energy lost| = |kinetic energy gained|

For the cylinder: mgL sin(30^{o}) = ½mv^{2}+ ¼mr^{2}(v^{2}/r^{2}) = ¾mv^{2}= ¾m 2aL,

a = (2/3) g sin(30^{o}),

since v^{2}- v_{0}^{2}= 2aL for motion with constant acceleration.

For the sphere: MgL sin(θ) = ½Mv^{2}+ (1/5)Mr^{2}(v^{2}/r^{2}) = (7/10)Mv^{2}= (7/5)MaL,

a = (5/7) g sin(θ).

The acceleration does not depend on the radii and masses of the cylinder and the sphere,

Equating (2/3) g sin(30^{o}) = (5/7) g sin(θ), sin(θ)= (14/15)sin(30^{o}),

θ = sin^{-1}(7/15) = 27.8^{o}.

An inclined plane of length L makes an angle θ > 0^{o} with respect
to the horizontal. A very thin ring with radius R rolls down this plane.
It reaches the bottom in time Δt_{ring}.

(a) If a uniform density
disk or sphere with the same radius R roll down this plane, at what time do they
reach the bottom? Find Δt_{disk}/Δt_{ring} and Δt_{sphere}/Δt_{ring}.

(b) If all three object should reach the bottom at the same time Δt_{ring},
one needs to vary the angle θ for the disk and the sphere. Find sinθ_{disk}/sinθ
and sinθ_{sphere}/sinθ.

Solution:

- Concepts:

Rigid-body motion, energy conservation, kinematics - Reasoning:

Energy conservation: mgLsinθ = ½mv^{2}+ ½I(v^{2}/r^{2})

Kinematics: v^{2 }= 2aL, a = ½gsinθ, L = ½at^{2}, t = (2L/a)^{½} - Details of the calculation:

(a) Ring:

Energy conservation: mgLsinθ = ½mv^{2}+ ½mr^{2}(v^{2}/r^{2}), gLsinθ = v^{2}.

Kinematics: v^{2 }= 2aL, a = ½gsinθ, L = ½at^{2}, t = (2L/a)^{½}= (4L/(gsinθ))^{½}.

Disk:

Energy conservation: mgLsinθ = ½mv^{2}+ ½(½mr^{2})(v^{2}/r^{2}), gLsinθ = ¾v^{2}.

Kinematics: v^{2 }= 2aL, a = (2/3)gsinθ, L = ½at^{2}, t = (2L/a)^{½}= (3L/(gsinθ))^{½}.

Sphere:

Energy conservation: mgLsinθ = ½mv^{2}+ ½(2/5)mr^{2}(v^{2}/r^{2}), gLsinθ = (7/10)v^{2}.

Kinematics: v^{2 }= 2aL, a = (5/7)gsinθ, L = ½at^{2}, t = (2L/a)^{½}= (14/(5gsinθ))^{½}.

Δt_{disk}/Δt_{ring}= (3/4)^{½}.

Δt_{sphere}/Δt_{ring}= (7/10)^{½}.

(b) The same acceleration will produce the same time Δt.

Disk: (2/3)gsinθ_{disk}= ½gsinθ, sinθ_{disk}/sinθ = ¾.

Sphere: (5/7)gsinθ_{disk}= ½gsinθ, sinθ_{disk}/sinθ = 7/10.

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:

Energy conservation - Reasoning:

No friction --> 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 dv_{x}/dt = 0, the acceleration changes sign. The force of constraint now must pull on the lower cylinder to keep the cylinders in contact. 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 dv_{x}/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. - Details of the calculation:
Let M be the mass of each cylinder and R the radius. Assume the cylinders stay in contact.

Energy conservation:

Mg(2R – 2Rsinθ) = ½ M(v_{x}^{2}+ v_{y}^{2})

The upper cylinder is moving down and the lower cylinder is moving towards the right.

v_{x}^{2}+ v_{y}^{2}= 4Rg(1 – sinθ)

X = 2Rcosθ, Y = 2Rsinθ,

v_{x}= dX/dt, v_{y}= dY/dt, v_{x}^{2}+ v_{y}^{2}= 4R^{2}(dθ/dt)^{2}(dθ/dt)^{2}= (g/R)(1 – sinθ)

Energy conservation yields dθ/dt in terms of the angle θ. Since v_{x}is proportional to dθ/dt, we can now find v_{x}and dv_{x}/dt in terms of θ, and therefore find the angle θ for which dv_{x}/dt = 0.

v_{x}= -2R sinθ dθ/dt, v_{x}^{2}= 4Rg(sin^{2}θ - sin^{3}θ)

dv_{x}^{2}/dt = (dv_{x}^{2}/dθ)(dθ/dt) = 2v_{x}(dv_{x}/dt)

dv_{x}^{2}/dt = 4Rg(2sinθ – 3sin^{2}θ) (dθ/dt)

dv_{x}/dt = dv_{x}^{2}/dt = 0 if 2sinθ – 3sin^{2}θ = 0.

sinθ = ⅔, dθ/dt = -(g/3R)^{½}, v_{x}= (4/3)(gR/3)^{½}.

The maximum speed of the bottom cylinder is (4/3)(gR/3)^{½}.