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Centripetal Force/ conservation of momentum/ etc

  • Thread starter FestiveF
  • Start date

FestiveF

Hi guys! I have solved some physics problems and was wondering if anyone would check over my work to reassure me of my answers? That would be great!

1) A pilot having a mass of 75 kg flies a plane in a vertical loop. At the top of the loop, where the plane is completely upside down, the pilot hangs freely in the seat and doesn't push against the seat belt. The airspeed indicator reads 120 m/s. What is the radius, in meters, of the plane's loop?

To solve this, i used the equation g= v^2/r. I am not 100% sure that this is the correct equation, but i was baffled as to what else would work. 9.8=120^2/ r
r= 120^2/9.8
r= 1469.39 m

2) In the process of running a ballistics test in a police laboratory, an officer fires a bullet having a mass of 6 g at a speed of 350 m/s into a container that slides .3 m across the table top. Assume the direction of the bullet is positive. What average force stops the bullet?

I began with the equation for momentum. p=mv
p=(.06kg)(350m/s)
p= (21)
Next, i think I need to find the time using v=d/t
350m/s=(.3m)/t
t= 8.57 x 10^-4
to find the average force, I then used F= v/t
F=350 (m/s)/8.57 x 10^-4 (s)
F= 408401.4 N

3) A baseball having a mass of .145 kg is pitched at 42 m/s. The batter hits the ball horizontally toward the pitcher at 58 m/s. If the ball and bat were in contact for 4.6 x 10^-4 s, what was the average force during contact? Assume the direction of the pitch is positive.

When the ball is first put into motion, it is going at 42 m/s. It is then returned at -52 m/s. (Negative due to the direction in comparision versus the original pitch).
(change in)p= m(change in)v
p= (.145 kg)(-100m/s)
p= -14.5
so..... F= (change in)p/(change in)t
F= (-14.5)/(4.6 x 10^-4)
F= -31521.74

4) A ball of mass .18 kg is placed on a compressed spring of the floor. the spring exerts an average force of 3.2 N over a distance of 19 cm as it propels the ball upward. How high, in cm, with the ball travel above the spring at the highest point of release?

I used W= Fd (J) --> 9change in) PE= mgh to solve this
W= 93.2 N) (.19m)= .608 J
W/(mg)=h
.608 J/ (.18*9.8)=h
h= .34 m or 34 cm

5) A rock of mass .4 kg is placed in the pouch of a sling shot. An average force of 8.2 N is used to pull the pouch and rock back a total of 43 cm. The rock is shot downward from a bridge that is located 18 m above a river. What will be the velocity or the rock, in m/s, an instant before it hits the water?

First you have to find the total energy of the system, correct? I used the equations W=FD and PE=mgh
W= (8.2 N)(.43 m)
W=3.526 J

PE=(.4 kg)(9.8)(18m)
PE=70.56 J

total= 3.526 J+ 70.56 J= 74.086 J

Then you can plug this energy into KE= .5 (mv^2)
v= (square root) 74.086 J/.5(.4)
v= 19.25 m/s

6) A circular Pyrex watch glass of 10 cm in diameter at 21 *C is heated to 510*C. In cm, what dimensional chage will be found in the circumference of the glass?

OK...the orginal circumference can be found by C= 2(pi)r
C=2(pi)(.05 m)
C= .314 m
Next I use the following:
(coefficient of linear expansion)= (change in) L/ Li (change in)t
3 x 10^-6=L / (.314)(489)
L= 4.52 x 10^-4 m or 4.52 x 10^-6 cm

Im really not sure if these are correct- any help/comments/suggestions will be greatly appreciated! i am trying to expand my knowledge in physics as much as I can! Thank you so much for your time!!!
 

Chi Meson

Science Advisor
Homework Helper
1,767
10
I'll do the first one:

You are correct, but you'll wnat to know why. In circular motion, you want to know what force (or combination of forces) cause the centripetal acceleration.

Cnetripetal force must always be supplied by a recognizable force such as tension, gravity, electromagnetic, or normal forces.

If the pilot is not straining against his (or her) straps and the seat is not pushing into him, then he is under the influence of gravity alone. At that moment, at the top of the circle, gravity supplies centripetal force.
 

Chi Meson

Science Advisor
Homework Helper
1,767
10
Oh, BTW. Sig figs, please.
 

FZ+

1,550
2
I'll do number 2....

2) In the process of running a ballistics test in a police laboratory, an officer fires a bullet having a mass of 6 g at a speed of 350 m/s into a container that slides .3 m across the table top. Assume the direction of the bullet is positive. What average force stops the bullet?
I began with the equation for momentum. p=mv
I would have thought energy considerations is the best for this one. And how come you didn't actually use this in your calculations?

p=(.06kg)(350m/s)
p= (21)

Oopsy, factor of 10 out. 6g is 0.006 kg...

Next, i think I need to find the time using v=d/t
350m/s=(.3m)/t
t= 8.57 x 10^-4

Ah... this is your big mistake. To use that formula for time, you are assuming the velocity is constant - but obviously it isn't.

to find the average force, I then used F= v/t
F=350 (m/s)/8.57 x 10^-4 (s)
F= 408401.4 N


Second big mistake - change in velocity/time is acceleration, not force.


Let's try a different method.

The key assumption to make is that the acceleration is more or less constant.

This lets us use a number of equations. The most useful is:

v^2 = u^2 + 2 * a * s

0 = 350^2 + 2 * a * 0.3
a = 204167 m/s^2

Kinda crazy, no?

But now this is the acceleration... F = ma... So multiply by your mass in kg to get a much more reasonable number in Newtons.
 
24
0
centripetal force/conservation of momentum/etc

Regarding number 4:

There are initially two forces acting on the ball: gravity and the spring. The net force is the force of the spring, less weight = (3.2 N)-(9.8*0.18) = 1.4 N

The work done on the ball and therefore its kinetic energy at the point of release is
W = Fd = (1.4 N) * (0.19 m) = 0.27 J

This is eventually transformed into gravitational potential energy at the top of the ball’s rise
0.27 J = mgh = (0.18 kg)(9.8 N/m)h
h = 0.15 m = 15 cm above the point of release.
 
24
0
Regarding number 6

It looks as though you know the coefficient of linear expansion is 3 X 10^(-6) /C. Sounds about right to me: I’ll take your word for it. However, it is linear not area.

If the final temp. is Tfinal and the initial temp is Tinitial, then
3 X 10^(-6)(Tfinal-Tinitial) = (change in length)/(initial length)

I believe your are being asked to solve for change in length.
 

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