Problem 1:
Two telephone poles are separated by 40 m and connected by a massless wire. A bird of mass 0.5 kg lands on the wire midway between the poles, causing the wire to sag 2.0 m below horizontal. What is the tension in the wire?
Solution:
 | Concepts, principles, relations
that apply to the problem: Static equilibrium: Ftotal = ttotal = 0 |
| Why do they apply? All parts of the system are at rest. |
| How do they apply? 2Fy = mg, Fy = 9.8*0.25 N = 2.45 N. sinq = 2/(4+400)1/2. F = Fy/sinq = 24.6 N is the tension in the wire. |
| Details of the calculation: None |
Problem 2:
A mass m starting from rest slides without friction in the field of gravity from the top of a vertical diameter d of a circle, A, along a straight wire connecting A and a point on the circumference, B. The wire makes an angle q with the diameter, as shown in the figure. Does the time it takes to slide from A to B depend on q? Justify your answer.
Solution:
| Concepts, principles, relations
that apply to the problem: Kinematics |
| Why do they apply? We are asked for the time it takes for an object to travel a given distance when it starts from rest and the acceleration is constant. |
| How do they apply? The distance x the mass travels is found from x/2 = (d/2)cosq.
The acceleration of the
mass is ax = gcosq. |
| Details of the calculation: None |
Problem 3:
A ball is dropped down an elevator shaft. The elevator has an upward speed V. The instant the ball is dropped, the top of the elevator is below it by a distance h. How high will the ball rebound?
Solution:
| Concepts, principles, relations
that apply to the problem: Conservation of mechanical energy, frame transformations |
| Why do they apply? Compared to the ball the elevator is “infinitely” massive. The frame of the elevator is an inertial frame. In the frame of the elevator, a dropped ball makes an elastic collision with the floor and bounces back. Its speed just before the collision equals its speed just after the collision. |
| How do they apply? In the frame of the elevator, the ball has an initial speed V. It will contact the floor after a time interval Dt. We have VDt + (1/2)gDt2 = h, Dt > 0. The speed of the ball after the time interval Dt is V + gDt. In the frame of the elevator the speed of the ball just after the bounce is v = V + gDt, and the ball is moving upward. In the frame of the ground, the speed of the ball just after the bounce is v = 2V + gDt and the ball is moving upward. Its kinetic energy is (1/2)mv2. It will rise a distance Dh = (1/2)v2/g above the contact point, which is at h - VDt below the point from which the ball was dropped, since in the time interval Dt the elevator as risen a distance VDt. In the frame of the ground the ball will therefore rise a distance h' = Dh - h + VDt above the point from which it was dropped. |
| Details of the calculation: Dt = ((V2+2gh)1/2 - V)/g v = V + (V2+2gh)1/2 Dh = (V2+gh+V(V2+2gh)1/2)/g VDt = (V(V2+2gh)1/2 - V2)/g h' = (V2+gh+V(V2+2gh)1/2)/g + (V(V2+2gh)1/2-V2)/g - h = 2V(V2+2gh)1/2)/g In the frame of the ground, the ball rebounds to a height h' = 2V(V2+2gh)1/2)/g above the point from which it was dropped. |
Problem 4:
A brave physics student (an undergraduate, of course)
climbs aboard a high powered merry-go-round and goes to the center, at r = 0.
At time t = 0, the platform starts from rest (W
= 0) and begins to spin about its vertical axis with constant angular
acceleration a. Also at time t = 0,
the student begins to crawl radially outward at constant speed u, relative to
the platform.
Assuming the student does not slip, find the student’s acceleration in the
inertial frame of an outside observer.
Solution:
| Concepts, principles, relations that apply to the
problem: Motion viewed from an inertial and a non-inertial frame |
| Why do they apply? In non-inertial frames fictitious forces appear. Consider a particle moving with velocity v in a reference frame K which moves with velocity V(t) relative to the inertial frame K0 and rotates with angular velocity W(t). The equations of motion are in the non-inertial frame are mdv/dt = Finertial - mdV/dt + mr ´ dW/dt - 2mW ´ v - mW ´ (W ´ r). Here -mdV/dt = fictitious force due to acceleration of frame mr ´ dW/dt = fictitious force due to non-uniform rotation of frame -2mW ´ v = Coriolis force -mW ´ (W ´ r) = Centifugal force In the inertial frame we have Finertial = mdv/dt + mdV/dt - mr ´ dW/dt + 2mW ´ v + mW ´ (W ´ r). |
| How do they apply? For our student dv/dt = dV/dt = 0. Finertial = -mr ´ dW/dt - 2m v ´ W + mW ´ (W ´ r). In the inertial frame Ftangential = (mra + 2mvat), Frad = -m(at)2r, Ftangential = 3mra, Frad = -m(at)2r. In the inertial frame the acceleration is therefore given by atangential = 3ra, arad = -(at)2r Assume that the direction of W is the z-direction. Then atangential = 3ra  and
arad
= -(at)2r. |
| Details of the calculation: Transforming from polar to Cartesian coordinates we have  =
cosf i + sinf
j,
= -sinf
i + cosf j.Therefore ax = -(at)2cosf - 3rasinf, ay = -(at)2sinf + 3racosf. r = vt, f = (1/2)at2. ax = -(at)2cos((1/2)at2) - 3vtasin((1/2)at2), ay = -(at)2sin((1/2)aat2) + 3vtacosf((1/2)at2). |
Problem 5:
A trunk weighing 500N is to be pushed up a rough incline by an applied horizontal force F. The incline makes an angle of 40o with the horizontal. If a force of 1000N is sufficient to move the trunk at a constant velocity of 0.200m/s, what is the coefficient of kinetic friction between the trunk and the incline?
Solution:
| Concepts, principles, relations
that apply to the problem: Newton’s 2nd law, the force of friction, the normal force, the gravitational force |
| Why do they apply? The trunk is moving in an inertial frame with constant velocity. It is not accelerating, Newton’s 2nd law requires that the net force is zero. In addition to the applied force, the force of friction, the normal force, and the gravitational force are acting on the trunk |
| How do they apply? The vector sum of all the forces acting on the trunk must be zero. |
| Details of the calculation: Choose the axes of your coordinate system as shown. The x- and y-components of the net force must be zero. x-component: f + mgsin(40o) - Facos(40o) = 0. Solve for f. f = 1000N cos(40o)-500Nsin(40o) = 444.65N. y-component: N – mgcos(40o) – Fasin(40o) = 0. Solve for N. N = mgcos(40o) + Fasin(40o) = 1025.81N. Use f = mN to find the coefficient m. m = f/N = 0.43 |
Problem 6:
Two blocks of mass M and 3M are placed on a horizontal frictionless surface.
A light spring is attached to one of them, and the blocks are pushed together
with the spring between them. A cord holding them together is burned, after
which the block of mass 3M moves to the right with a speed of 2m/s.
(a) What is the speed of the block of mass M?
(b) Find the original elastic potential energy in the spring if M = 0.35kg.
Solution:
|
Concepts, principles, relations
that apply to the problem: |
|
Why do they apply? |
|
How do they apply? |
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Details of the calculation: |
Problem 7:
The turns at the Daytona Speedway have a radius of 300 m
and are banked at about 30o.
(a) At what speed does the banked curve provide exactly the centripetal
acceleration required to move the car around the curve?
(b) Ignoring factors other than gravity, what is the maximum speed of the car
without sliding, if the coefficient of friction between the car and tires is
0.7?
(c) What additional forces on the race car allow it to exceed this speed?
Solution:
| Concepts, principles, relations
that apply to the problem: Uniform circular motion, Fc = mv2/r |
| Why do they apply? In part (a) the horizontal component of the normal force must provide all the centripetal acceleration and in part (b) the frictional force must provide part of the centripetal acceleration. |
| How do they apply? (a) A free-body diagram of the car on the track is shown below.
(b) Assume the car is traveling with speed v' and f = fmax.
Then |
| Details of the calculation: None |
Problem 8:
A massless string of length L has a ball of mass m attached to one end and the other end is fixed. The ball is launched so that it moves in a vertical circle in a gravitational field, with an initial velocity v0 downward. What is the minimum value for v0 if the ball is to rotate around on a circle of radius L?
Solution:
| Concepts, principles, relations
that apply to the problem: Energy conservation, centripetal acceleration |
| Why do they apply? Place the origin of the coordinate system at the center of the circle and let the y-axis point upwards. For y > 0 the centripetal acceleration needed to keep the ball moving on the circle is partly provided by gravity and the tension in the string decreases. When the tension in the string vanishes then the ball behaves like a free particle in a gravitational field. The minimum value for v0 is the one for which the tension vanishes when the ball is on the top of the circle. |
| How do they apply? Energy conservation: E = (1/2)mv02 = (1/2)mv2 + mg Lsinf. Here f is the usual polar angle. v2/L = v02/L - 2gsinqf is the centripetal acceleration. The tension in the string vanishes when v2/L = gsinf, when all the centripetal acceleration is due to gravity. For the ball is to rotate around on a circle of radius L we need v2/L ³ g when f = 90o. We need v02/L ³ 3g, v0 ³ (3gL)1/2. The minimum value of v0 is (3gL)1/2. |
| Details of the calculation: None |
Problem 9:
A remote spherical planet of radius c is discovered to have
an empty core! A small mining robot of mass m finds that the planet has an
empty spherical void located at its center of radius d. The density of the
massive portion of the planet (which extends from a radius of d out to c) is
r.
(a) What is the gravitational force on the robot when it is located at a radius
less than d?
(b) In which direction is the force on the robot oriented when it is located at
a radius slightly greater than d?
(c) Where within the massive portion of the planet would the magnitude of the
force on the robot be the least?
(d) What is the force on the robot when it is located at a radius r between d
and c?
(Be sure to express your answer in terms of the variables defined in this
problem.)
Solution:
| Concepts, principles, relations
that apply to the problem: Newton's law of gravitation, Gauss' law |
| Why do they apply? Gauss’ law: (Ñ×Fg/m) = -4pGr, for a spherical mass distribution. 4pr2(Fg/m) = 4pGMinside, (Fg/m) = GMinside/r2. Here Fg/m is the gravitational force per unit mass. The direction of Fg is towards the center of the planet. |
| How do they apply? (a) F = 0 (b) The direction of F is towards the center of the planet. (c) F(r) = GmMinside/r2 = Gmr4p(r3 – d3)/(3r2) = (4/3)prGm(r - d3/r2). F(r)min = 0 at r = d. (d) F(r) = (4/3)prGm(r - d3/r2) |
| Details of the calculation: None |
Problem 10:
A free particle with mass m is subjected to a
time-dependent force F(t) = F0 exp(-t/t),
where F0 and t are constants.
(a) Find the velocity v(t) of the particle as a function of time, given that v
= 0 at time t = 0.
(b) What is the total work done on the particle by the force?
Solution:
| Concepts, principles, relations
that apply to the problem: Newton’s 2nd law, the work-kinetic energy theorem |
| Why do they apply? Given the acceleration and v(0) we can find v(t) and the kinetic energy at time t. The net work done on the object is equal to the change in the kinetic energy of the object. |
| How do they apply? (a) F = ma = mdv/dt = F0 exp(-t/t), v(t) = (F0t /m)(1-exp(-t/t)). (b) As t à infinity, v à (F0t /m). Total work done by the force: W = ½mvinf2 = ½(F0t)2/m. |
| Details of the calculation: None |
Problem 11:
Consider a particle moving along the x axis under the influence of the potential
U(x) = (1/2)k(x/a)2 - (1/3)k(x/a)3
where k and a are constants.
(a) Plot the potential.
(b) Find the equilibrium point(s).
(c) Determine whether the equilibrium point(s) are stable or unstable.
Solution:
| Concepts, principles, relations
that apply to the problem: Dynamic equilibrium |
| Why do they apply? We are asked to find the equilibrium point(s), where F = 0. An equilibrium point is stable if for small displacements from the equilibrium point the force points towards the equilibrium point. |
| How do they apply? (a) Let k = 1.
|
| Details of the calculation: None |
