Radial force on a particle with circular motion

In summary, a particle of mass 0.014 kg traveling on a radius of 0.9m at a rotational velocity of 36+5*t rad.s-1 will complete 36.69 revolutions in 4.8 seconds. The radial force on the particle at 4.8 seconds is 45.36 N, and the absolute magnitude of force is 45.36 N. The angular acceleration is 5 rad/s^2 and the linear acceleration is 4.5 m/s. The direction of the radial force may be considered in comparison to its direction at t=0.
  • #1
Lee333
4
0
A particle of mass 0.014 kg is traveling on a radius 0.9m at a rotational velocity of 36+5*t rad.s-1.


How many revolutions does the particle go thorugh in 4.8 seconds? [revs]


What is the radial force on the particle at 4.8 seconds? [N] (Hint: consider directions)
(question I'm stuck on)


What is the absolute magnitude of force on the particle at 4.8 seconds? [N]



We know that the radial acceleration = v^2/r



To find the amount of revolutions I simply integrated the rotational velocity with respect to time, and found that the particle rotated 230.4 radians or 36.69 revs.

The second question is the question that I seem to be getting wrong. The way I tried to solve it was by firstly finding the rotational velocity after 4.8 seconds:
w = 36+5*4.8
= 60 rad/s

thus we know that the translational velocity is:
v= rw
= 60*0.9
= 54 m/s

we know that the radial acceleration, ra, can be defined by:
ra = v^2/r
= 54^2/0.9
= 3240

thus the radial force, Fr, is:
Fr = m*ra
=0.014*3240
=45.36 N

the above question i am getting wrong for some reason. any help would be appreciated




For the final question we know that the angular acceleration is 5 rad/s^2
this means the linear acceleration of the particle is 4.5 m/s.

we know the translational force will be Ft=m*a
Ft = 0.014*4.5
= 0.063

So we can work out the total magnitude of the force like so:
F_tot = sqrt(0.063^2+45.36^2)
= 45.36
 
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  • #2
hi lee, welcome to physicsforums :)
Your answer looks good to me. Maybe they also wanted the direction of the radial force at that time.
 
  • #3
Thanks for answering, I think its got something to do with the consider directions hint, but I am not sure what I've done wrong. I don't think they want the direction of the radial force because they didn't give a starting position for the particle, so we don't know the direction of the force.
 
  • #4
ah, that's true, they don't give the starting position. Then maybe you are meant to give the direction of the radial force, compared to its direction at t=0 ? This is a bit of a stretch, but I can't imagine what else they want for the answer...
 
  • #5
N


First, let's address the issue with the second question. You have correctly calculated the radial acceleration, but you have forgotten to take the direction into account. Remember, force is a vector quantity and has both magnitude and direction. In this case, the radial force is directed towards the center of the circular motion, so it should be a negative value. The correct answer for the radial force at 4.8 seconds would be -45.36 N.

For the final question, you have correctly calculated the magnitude of the force. Just remember to include the direction as well, which in this case would be towards the center of the circle. So the absolute magnitude of the force would be 45.36 N.

Overall, your approach and calculations are correct, just remember to include the direction of the force in your answers. Keep up the good work!
 

1. What is the radial force on a particle with circular motion?

The radial force on a particle with circular motion is the force that acts towards the center of the circular path of the particle. It is responsible for maintaining the particle's circular motion.

2. How is the magnitude of radial force calculated?

The magnitude of radial force is calculated using the formula F = m(v^2)/r, where m is the mass of the particle, v is its velocity, and r is the radius of the circular path.

3. What happens if there is no radial force acting on a particle with circular motion?

If there is no radial force acting on a particle with circular motion, the particle will continue to move in a straight line tangent to its circular path, rather than following the circular path.

4. How does the direction of the radial force change as the particle moves in its circular path?

The direction of the radial force changes continuously as the particle moves in its circular path. It always points towards the center of the circle, and its direction is perpendicular to the velocity of the particle at any given point.

5. Can the radial force affect the speed of the particle with circular motion?

Yes, the radial force can affect the speed of the particle with circular motion. If the radial force increases, the particle will accelerate and its speed will increase. Similarly, if the radial force decreases, the particle will decelerate and its speed will decrease.

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