course 201

These were the problems you sent out a while back for us to prepare for Test Two" "Phy 201 problems, Test 2

Problem Number 1

The energy required for a 780 kg mass to climb from the surface of the Earth to a point 6400 km 'above' the surface--i.e., at two Earth radii from the center--can be approximated by first averaging the field strength of the Earth at this point and at its surface.

• Given a surface field strength of 9.8 m/s^2, what is the field strength at two Earth radii from its center? What is the average of these two strengths?

Fgravsurface=9.8m/s^2(780 kg)=7644 N

Fgrav=GMm/r^2=(6.67x10^-11 Nm^2/kg^2)(6X10^24 kg)(780 kg)/2(6400000 m)^2=1904.3 N

7644 N+1904.3 N/2= 4774.2 N

• Using this average to approximate the actual average field strength determine the work necessary to climb from the surface to a point two Earth radii from its center.

Work=Fnet*`ds=4774.2 N (6400000 m)= 3.06x10^10 mN

??? is it correct to refer to it as ‘mN’ in this situation? I can’t remember how to distinguish the two usages

In this case you are calculating work, which is the product of average force and displacement. The unit of work could be expressed as N * m or m * N, since multiplication is commutative, but when it's work we generally write it as N * m = Joules.

The other usage is when we are finding torque, which is obtained by multiplying meters of moment arm by Newtons of force. The order of that definition is r X F, and we usually write torque as m * N. It is no appropriate to refer to a torque in Joules. The technical reason torque can't be referred to in Joules that torque is a vector quantity (direction by r.h. rule, r crossed with F), and work is a scalar quantity (just a magnitude, no direction). They are two completely different kinds of things, and the Joule is the unit for work.

Problem Number 2

A cart of mass 1.9 kg coasts 45 cm down an incline at 6 degrees with horizontal. Assume that frictional and other nongravitational forces parallel to the incline are negligible.

• What is the component of the cart's weight parallel to the incline?

Mg=9.8 m/s^2 (1.9 kg)=18.6 N

Mgx=18.6 N(cos 276)=1.94 N

• How much work does this force do as the cart rolls down the incline?

Work= Fnet*`ds=1.94 N(.45 m)=.875 J

• Using the definition of kinetic energy determine the velocity of the cart after coasting the 45 cm, assuming its initial velocity to be zero.

Work=`dKE

.875 J=.5(1.9 kg)(Vf^2)-0

Vf^2=.921 m^2/s^2

Vf=.96 m/s

good

• Using the definition of kinetic energy determine the velocity of the cart after coasting the 45 cm, assuming its initial velocity to be .34 m/s.

Work=`dKE

.875 J=.5(1.9 kg)(Vf^2)-.5(1.9 kg)(.34 m/s)

Vf^2= 1.04 m^2/s^2

Vf=1.02 m/s

good

Problem Number 3

A disk of negligible mass and radius 30 cm is constrained to rotate on a frictionless axis about its center. On the disk are mounted masses of 7 grams at a distance of 23.1 cm from the center, 24 grams data distance of 16.8 cm from the center and 47 grams at a distance of 9 cm from the center. A uniform force of .06983 Newtons is applied at the rim of the disk in a direction tangent to the disk.

• What will be the angular acceleration of the disk?

Tau=I*alpha

Tau= r*F= .3 m(.07 N)=.021 mN

I=sigma .5mr^2=.5(.007 kg)(.231 m)^2 + .5(.024 kg)(.168 m)^2 +.5(.047 kg)(.09 m)^2= 7.16 *10^-4 kgm^2

.021 mN =7.16 *10^-4 kgm^2 (omega)

alpha= 29.3 rad/sec^2

• If the force is applied for 4 seconds with the disk initially at rest, what angular velocity with the disk attain?

Taunet *`dt=`dI*omega

.021 mN (4 sec)=7.16 *10^-4 kgm^2 (omega final)-0

Omegafinal=1.173 *10^-6 rad/sec

.021 * 4 / (7.16 * 10^-4) = 1.2 * 10^2, not 1.2 * 10^-6

One way you might have gotten your incorrect result would be to calculate

.021 * 4 / 7.16 * 10^-4, which by order of operations would equal your result, rather than

.021 * 4 / (7.16 * 10^-4).

• What then will be the speed of each of the masses?

?? I am unsure of how to calculate this.

Each mass follows a circular path about the axis, at about 120 rad/s.

Each radian of angular motion corresponds to a distance on the arc equal to the radius.

So if you multiply 120 rad / s by the radius of the path followed by each mass, respectively, you get its speed.

This will allow you to calculate the KE of each mass; adding them gives you the total KE, as requested in the next step.

• What will be their total kinetic energy?

• Compare the total kinetic energy to the change in the quantity .5 I `omega^2.

You have I and omega, so you can independently calculate KE = 1/2 I omega^2. Your result should match the total KE you got in the previous step.

Problem Number 4

A cart of mass 350 grams is temporarily held stationary on an incline at 5.9 degrees to horizontal.

• A mass of 63.2 grams is attached to the cart by a string over a pulley located at the bottom of the incline; the mass hangs freely from the string.

• When the cart is released, what will be its acceleration in the direction down the incline?

Mga=9.8 m/s^2(.0632 kg)=.6194 N

Mgb=9.8 m/s^2 (.35 kg)=3.43 N

Mgbx=3.43 N cos(275.9)= .353 N

Fnet= Mga+Mgbx=.6194 N+.353 N= .972 N

Fnet=m*a

.972 N=(.0632 kg+.35 kg)*a

A=2.35 m/s^2

goot

• Sketch and label a reasonable facsimile of a scaled diagram showing the forces acting on the cart.

Problem Number 5

A gun fires a bullet of mass 9 grams out of a barrel 38 cm long. The gun is attached to a spring. From the recoil of the spring and the masses of the gun and the spring we determine that the gun recoiled with a total momentum of 2 kg m/s.

• With what velocity did the bullet exit the barrel?

Momentum=mvf-mvo

2 kgm/s=(.009 kg) Vf-0

Vf-222.2 m/s

• Assuming that the bullet accelerated uniformly from rest along the length of the barrel, how long did it take the bullet to accelerate from rest down the length of the barrel?

`ds=vf+v0/2*`dt

.38 m=222.2 m/s+0/2*`dt

`dt=.00342 seconds

• What was the average force exerted on the bullet as it accelerated along the length of the barrel?

Fnet*`dt=mvf-mv0

Fnet (.00342 sec)=(.009 kg)(222.2 m/s)-0

Fnet=584.7 N

• What average force would be felt by the individual holding the gun for the time the bullet accelerates along the length of the barrel?

?? I’m not sure which formula to use for this one

The force exerted by the bullet on the gun is equal and opposite to the force exerted by the gun on the bullet, which you calculated previously.

The gun is in contact with the person. Assuming that the gun pushes on the person and on the bullet for an identical time interval (which is not necessarily the case, since it would probably take the person a bit longer to bring the gun to rest) the force of the person on the gun would be equal to the force exerted by the gun on the bullet.