Force on a Rolling Sphere: 2.4x10^-3 N

  • Thread starter Thread starter John O' Meara
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
A uniform sphere with a radius of 5.0 cm and a mass of 2.0 kg rolls to rest over a distance of 15 m from an initial speed of 30 cm/s. The calculated stopping force is 2.4 x 10^-3 N based on angular velocity and torque. However, there is debate about the method used, with suggestions that it may be more appropriate to treat the problem as a work-energy scenario. The stopping force's application point is also questioned, as it may affect the calculations if applied at the sphere's edge versus its axis. The discussion highlights the importance of considering the force's point of application and the potential for friction to influence the outcome.
John O' Meara
Messages
325
Reaction score
0
A uniform sphere of radius 5.0cm and 2.0kg mass is rolling along level ground at a speed of 30cm/s. It rolls to rest in a distance of 15m. How large a stopping force acted on it.
15/(2*PI*.005)=47.75rev = 300 rad
15m/0.3m/s=50s => w=6 rad/s. w_avg = (6+0)/2=3rad/s, therefore (alpha)=-.06rad/s/s. Torque T=I*(alpha). I=2/5*M*a^2. Therefore T = -1.2*10^-4 N.m; rxF=T, => F=2.4*10^-3N. Is this correct? Am I on the right track? Thanks.
 
Physics news on Phys.org
15/(2*PI*.005)=47.75rev = 300 radshouldn't 0.005 be 0.05.
 
but looks like you did use 0.05 to get the answer you have, just have a typo.
 
I used the average angular velocity in this calculation to get the right answer, but why use the average value and not the value given?Thanks
 
John O' Meara said:
I used the average angular velocity in this calculation to get the right answer, but why use the average value and not the value given?Thanks
\omega_f^2 = 2(\alpha)(\theta)
where \omega = 6 and \theta =300
solve \alpha = .06
 
John O' Meara said:
A uniform sphere of radius 5.0cm and 2.0kg mass is rolling along level ground at a speed of 30cm/s. It rolls to rest in a distance of 15m. How large a stopping force acted on it.
15/(2*PI*.005)=47.75rev = 300 rad
15m/0.3m/s=50s => w=6 rad/s. w_avg = (6+0)/2=3rad/s, therefore (alpha)=-.06rad/s/s. Torque T=I*(alpha). I=2/5*M*a^2. Therefore T = -1.2*10^-4 N.m; rxF=T, => F=2.4*10^-3N. Is this correct? Am I on the right track? Thanks.
I doubt that this is the correct approach to this problem. Were you given any information about where this stopping force is applied? Why do you assume it was applied at the edge of the ball? Could it be that this is a work/energy problem?
 
v=.3m/s ,s=15m. .5*m*v^2 + .5*I*w^2 = F*s, where I=2/5*m*(.05)^2
threrfore: .09 + .036 =F*15 => F=8.4*10^-3
 
John O' Meara said:
v=.3m/s ,s=15m. .5*m*v^2 + .5*I*w^2 = F*s, where I=2/5*m*(.05)^2
threrfore: .09 + .036 =F*15 => F=8.4*10^-3
I did not check your computatiuon, but that is conceptually correct IF the force is appied at the axis of rotation. If the force were applied at the top of the ball, the work would be the same, but the force would be only half as much because the distance over which that force is appied would be the distance a point on the vertical equator of the sphere moves, twice as far as the center of the ball moves. All this assumes the ball keeps rolling without slipping. If it slipped, friction would do some of the work.
 
Back
Top