Minimum linear velocity attained by sphere

[utkarshakash]

Homework Statement

A sphere of mass M and radius R is moving on a rough fixed surface, having co-efficient of friction μ, with a velocity v towards right and angular velocity ω clockwise. It will attain a minimum linear velocity at time (take v>ωR)

The Attempt at a Solution

Since v>ωR the sphere rolls with slipping. So frictional force will act in the backward direction. Using the equation \int \tau dt = \int dL where τ=μmgR.

\mu mgRt= \frac{2}{5} mR^2 (\omega &#039; - \omega) \\<br /> <br /> \mu mg = m \frac{dv}{dt} \\<br /> <br /> \mu gt = (v&#039; - v)<br />

Using the relation v'=ω'R and solving the above two equations I get
t= 2(v-ωR)/3μg. But the correct answer has 7 in the denominator.

 

[mfb]

\mu mg = m \frac{dv}{dt}

There is a minus sign missing.

I don't see why this point is called "minimum linear velocity" - it is the point where the sphere stops slipping.

 

[utkarshakash]

There is a minus sign missing.

I don't see why this point is called "minimum linear velocity" - it is the point where the sphere stops slipping.

Why there should be a minus sign? Since friction acts backwards, so is the acceleration. Hence both minus cancels out. Are you trying to say that since velocity is decreasing that's why dv/dt should carry a '-' with it?

 

[mfb]

Acceleration is backwards, but as you write your equation both sides are positive, so v increases. You can use a negative v everywhere, but then things get really confusing and the initial v>ωR does not work.