What Forces Act on a Mass Sliding Inside a Hoop?

  • Thread starter Thread starter Staerke
  • Start date Start date
  • Tags Tags
    Mass Rotating
AI Thread Summary
A mass of 0.52 kg slides inside a frictionless hoop with a radius of 1.10 m and a speed of 4.25 m/s at the top. The total energy at the top is calculated to be 15.9189 J, which is used to determine the speed at a 44° angle, yielding a velocity of 7.44503 m/s. The centripetal force is initially calculated as 26.2026 N, but adjustments for the gravitational component are necessary. The net force exerted on the mass includes both the centripetal force and the gravitational force's radial component. Properly accounting for these forces is essential to find the correct force with which the mass pushes on the hoop.
Staerke
Messages
12
Reaction score
0

Homework Statement


A mass M of 5.20E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of 4.25 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 44.0°.


Homework Equations


E = Pe + Ke
Pe = m * g * h
Ke = 1/2 * m * v^2
F=mv^2/r


The Attempt at a Solution



Alright so energy at the top = Pe + Ke
So m*g*h + 1/2*m*v^2 = 11.2226 + 4.69625 = 15.9189

The energy in the bottom is going to be equal to the energy at the top
The height at the bottom is R-Rcos(44) = .295511
15.9189 = .52*9.81*.29511 + .5*.52*v^2
V=7.44503

Centripital force = mv^2/r, so .52*7.44503^2/1.1 = 26.2026 N

But this is wrong. Now I thought it might be reduced due to gravity.
so F=26.2026-m*g*cos(44) = 22.5331
Neither answer works.
Help?
 
Physics news on Phys.org
Nobody? :(
 
Staerke said:

Homework Statement


A mass M of 5.20E-1 kg slides inside a hoop of radius R=1.10 m with negligible friction. When M is at the top, it has a speed of 4.25 m/s. Calculate the size of the force with which the M pushes on the hoop when M is at an angle of 44.0°.

Homework Equations


E = Pe + Ke
Pe = m * g * h
Ke = 1/2 * m * v^2
F=mv^2/r

The Attempt at a Solution



Alright so energy at the top = Pe + Ke
So m*g*h + 1/2*m*v^2 = 11.2226 + 4.69625 = 15.9189

The energy in the bottom is going to be equal to the energy at the top
The height at the bottom is R-Rcos(44) = .295511
15.9189 = .52*9.81*.29511 + .5*.52*v^2
V=7.44503

Centripetal force = mv^2/r, so .52*7.44503^2/1.1 = 26.2026 N

But this is wrong. Now I thought it might be reduced due to gravity.
so F=26.2026-m*g*cos(44) = 22.5331
Neither answer works.
Help?
Sorry that nobody responded sooner.

What is the angle 44.0° in relation to.

Your answer looks right for centripetal force, assuming the velocity is right.

The radial component of the net force exerted on M is equal to Mv2/r, the centripetal force. However, there are two contributions to this force.
One is the force exerted on M by the hoop.

The other is gravity. There a component of the gravitational force which is in the radial direction. This is what you haven't taken into account.​
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top