Can someone check my work? (centripetal/friction forces)

  • Thread starter dnt
  • Start date
  • #1
dnt
238
0
I think these are correct but can someone verify? plus i have a few questions as well.

i know its long but any help is much appreciated! thanks.

1) a .40 kg ball is attached to the end of a horizontal cord and is rotated in a circle (r=1.3 m) on a frictionless surface. if the cord will break when the tension exceeds 60 N what is the max speed? and how would your answer be affected if there was friction?

my answer: i set F(c) = F(tension) and solved for v = 13.96 m/s. if there was friction my answer would go up.

side question: how do you know what direction tension always points in since the string/cord goes in both directions? and is it true that Fc always points inward? im having trouble learning the direction of the forces, particularly centripetal and tension.

---------------------------------------------

2) what is the max speed with which a 1050 kg car can round a turn (r = 70m) on a flat road if the coefficient of friction is .80? is this result independent of the mass of the car?

my answer: i set F(fr) = F(c) and solved for v = 23.43 m/s. since the masses cancelled out on both sides, the answer is independent of the mass of the car.

---------------------------------------------

3) a ball on the end of a string is revolved at a uniform rate in a vertical cirlce (r = .85m). if its speed is 4.15 m/s and the mass is .3 kg, what is the tension at the top and at the bottom?

my answer: for the top, i had F(c) pointing up, F(tension) and F(w) pointing down, therefore F(c) = F(tension) + (Fw). solving for tension i got 3.1 N.

on the bottom, F(c) now points down and F(tension) points up so F(c) + F(w) = F(tension). again solving for tension i got 9.0 N.

again, the direction of forces is confusing me. am i close on these?

---------------------------------------------

4) how large must the coefficient of friction be between the tires and the road if a car is to round a level curve of r = 85 m at a speed of 95 km/h.

my answer: first i converted the speed to 29.39 m/s, then set F(fr) = F(c) and solved for the coefficient which i got .84.

---------------------------------------------

5) at what minimum speed must a roller coaster be traveling when upside down at the top if the passengers are not to fall out (r = 8.6 m).

my answer: I set F(w) = F(c) since F(w) points down and F(c) points up and solved for v = 9.18 m/s

side question: i know that in basically all forces questions, net force = ma. and in these the forces should be cancelling out (ie, net acceleration = 0 and therefore you can set the two forces equal to each other becaues you can add one to the other side). but when, in circular questions, can you not assume a=0? how does that change things?

---------------------------------------------

6) a 1200 kg car rounds a curve of r=70m banked at a angle of 12 degrees. if the car is traveling at 90 km/h will a friction force be required? if so how much and in what direction?

my answer: this one i havent fully solved yet. again i cannot get the direction of the forces. on my diagram (car on a slant), i have F(w) directly down, F(N) pointing perpendicular upwards, and then F(c) pointing out (or "up" the slant) and F(fr) pointing oppositely inward (or "down" the slant). both F(c) and F(fr) are parallel to the slant.

now i know that F(N) will be equal to F(w)cos12 and F(fr) = F(w)sin12.

does this mean i can set F(w)sin12 = Fa? i get a coefficient of .911. is that right? does it mean you do need friction? and again the direction would be inward?

thanks for the help.
 

Answers and Replies

  • #2
dnt
238
0
anyone? just a little help would be very appreciated! thanks.
 
  • #3
dnt
238
0
bump. can someone at least help me with the direction of forces (centripetal, friction and tension). thanks.
 
  • #4
dnt
238
0
one more bump, hoping to get a little help :)

(or if someone could tell me why im not getting any help so i know...did i break a rule? am i not posting correctly? :confused: )
 
  • #5
Astronuc
Staff Emeritus
Science Advisor
19,798
3,245
To answer the question about tension, centripetal and centrifugal force, see - http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html

http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html#circ

http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html#strmas

1. 13.96 m/s is correct. If the ball is rolling, then one applies static friction which will help resist the ball wanting to roll away.

2. v = 23.43 m/s is correct and since mass divides out, then the velocity is independent of mass.

3. refer to hyperphysics links above. When the ball is at the top of the arc, the 'centrifugal' force operates opposite gravity, and at the bottom, the centrifigual force operates with gravity.

4. 95 km/h = 95000 m/6300 s = 26.4 m/s, but you have the right idea.

5. v = 9.18 m/s is correct

6. Banking allows one to use weight, or rather the component parallel with the banked surface in addition to the friction to maintain a vehicle on a curve.
 
  • #6
dnt
238
0
  • #7
Astronuc
Staff Emeritus
Science Advisor
19,798
3,245
I think the best explanation of centripetal and centrifugal force is found here - http://hyperphysics.phy-astr.gsu.edu/hbase/corf.html#cent

Whereas the centripetal force is seen as a force which must be applied by an external agent to force an object to move in a curved path, the centrifugal and coriolis forces are "effective forces" which are invoked to explain the behavior of objects from a frame of reference which is rotating.

The tension in a string provides the centripetal force on a mass (moving in an arc) which is attached to one end with the other end fixed.

Tension can be thought of as a pulling vs a pushing force.

F <---- ================= ----> F opposing forces pointing out from an element provide tension, so we call these tensile forces.

F ----> ================= <---- F opposing forces pointing into (in toward) an element provide compression, so we call these compressive forces.

If the forces is unbalanced, i.e. one of the opposing forces is greater than the other, then there is a net force and the element would have to be accelerating.
 

Related Threads on Can someone check my work? (centripetal/friction forces)

  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
1
Views
6K
  • Last Post
Replies
1
Views
9K
Replies
3
Views
2K
Replies
2
Views
619
Top