Sphere rolling with slipping on a movable platform

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SUMMARY

The discussion centers on the dynamics of a sphere rolling with slipping on a movable platform, specifically analyzing the final speed of the platform relative to water and the time of contact between the sphere and the platform. The derived formula for the time of contact is Δt = (L(7M + 2m)) / ((M + m)rω₀), highlighting the importance of both mass terms in the calculations. Participants emphasize the need for clarity in referencing points for angular momentum and the impact of kinetic friction on the sphere's motion.

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  • Newton's laws of motion
  • Concept of angular momentum
  • Understanding of kinetic friction
  • Basic rotational dynamics and moment of inertia
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Alkaid
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Homework Statement


A sphere (of radius r and mass m) rotating with angular velocity ω0 is lowered onto the edge of a floating platform of length L and mass M. The platform can move freely on water. The platform is rough and the sphere rolls all the way from one edge to the other edge of the platform with slipping except the very end of the other edge.

The question requires to find the final speed of the platform w.r.t. the water and also to show that the time of contact of the sphere and the platform is
\triangle t = \frac{L(7M+2m)}{(M+m)r\omega_0}

Homework Equations


Basically Newton's law of motion

The Attempt at a Solution


I want to know where did i get it wrong and I've done the following to find the time \triangle t but with the 2m term missing:

I first find the speed v_s of the sphere relative to the water:
Kinetic friction f_k = \mu mg = ma
Torque -f_k r = I \alpha
v_s = v_0 + at = at
angular speed \omega_s = w_0 + \alpha t
Denote the final speeds and final angular momentum of the sphere (at the other edge of the platform) as
v_f ,\omega_f respectively
\Rightarrow \omega_f = \omega_0 - \frac{mr}{I} v_f =\frac {v_f}{r} I is the moment of inertia of sphere
v_f = \frac{\omega_0}{\frac{1}{r}+\frac{mr}{I}}=\frac{2}{7}\omega_0
By Newton's third law,
mv_s = Mv_p where v_p is the velocity of the platform
\rightarrow v_p =\frac{m}{M} \frac{2}{7} \omega_0
The time will then be the distance divided by the average (since acceleration is constant) relative velocity of the sphere and the platform:
\triangle t = \frac{L}{\frac{1}{2}(1+\frac{m}{M})\frac{2}{7}\omega_0}=\frac{7ML}{(M+m)r\omega_0}
But i am missing the 2m term and I cannot figure out where i was wrong
 
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Alkaid said:
## \omega_f = \omega_0 - \frac{mr}{I} v_f ##
Sorry, but I don't understand how you get that equation. I assume this is from conservation of angular momentum, and your reference point is the centre of the sphere (or some fixed point in its path). If so, I would expect to see an M term but not an m term.
(One should always state the reference point/axis for considerations of angular momentum etc.)

Edit:
Alkaid said:
## \omega_f = ... =\frac {v_f}{r}##
That doesn't look right either. What about the linear motion of the platform?
By the way, I get a different answer from the one given.
 
Last edited:

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