Question

A point mass m start froms rest and slides down the surface of a frictionless solid sphere of radius r as shown in the figure. Measure angle from the vertical and potential energy from the top. Find:

(i) The angle at which the mass files off the sphere.

(ii) If there is friction between the mass and sphere, does the mass fly off at a greater or lesser angle than in part (a)?

• A. greater
• B. greater
• C. lesser
• D. lesser

Answer 1

The correct option is A

greater

Angle at which the mass flies off the sphere: A closer scrutiny of the force equation in the radial direction,

$Mgcos\theta -N=\frac{m{v}^2}{r}$ -------------------------(i)

is instructive to get the angle at which the mass flies off the sphere. As the mass slide down the sphere its speed increases (this is because $\frac{dv}{dt}$ is positive; or the potential energy of the mass decreases, and for mechanical energy to remain constant kinetic energy of the mass must increase). So the right hand side of the equation (i) $\frac{m{v}^2}{r}$, increases with increasing $\theta$, Also mg cos $\theta$ decreases as $\theta$ increases, (for 0 < $\theta$ < $\frac{\pi }{2}$, cos $\theta$ decreases as $\theta$ increases).

For the equality of the equation (i) to hold, N must decrease with increasing $\theta$. The angle at which N $\to$ 0, the mass loses contact with the sphere, it flies off it. Let this happen when $\theta \to \alpha$. At $\theta \to \alpha .N\to 0,v\to v_{\alpha }$. Under this limiting condition, equation (i) reduces to.

$\Rightarrow mgcos\alpha -0=\frac{m{v}_{alpha}^2}{r}$ -----------------------(ii)

Conservation of mechanical energy between $\theta =0$ and $\theta =\alpha$ gives,

$\Rightarrow \frac{1}{2}m{v}_{alpha}^2=mgr(1-cos\alpha )$ --------------------(iii)

From equation (ii) and (iii) $\Rightarrow \frac{\left(\frac{m{v}_{\alpha }^2}{r}\right)}{\left(\frac{1}{2}m{v}_{\alpha }^2\right)}=\frac{mgcos\alpha }{mgr(1-cos\alpha )}\Rightarrow \frac{\left(\frac{1}{r}\right)}{\left(\frac{1}{2}\right)}=\frac{cos\alpha }{r(1-cos\alpha )}$

$\Rightarrow 2=\frac{cos\alpha }{1-cos\alpha }\Rightarrow 2-2cos\alpha =cos\alpha$

$cos\alpha =\frac{2}{3}\Rightarrow \alpha =co{s}^{-1}\left(\frac{2}{3}\right)={48.2}^{\circ }$

If friction is present, the speed at $\theta =\alpha$ will be less than $v_{\alpha }$ defined in part (i), and mg cos $\alpha =\frac{m{v}^2}{r}$ will not be satisfied, in fact, then mg cos$\alpha >\frac{m{v}^2}{r}$ as $v<u_{\alpha }$ And, consequently, Newton's second law of motion will demand that.

$mgcos\alpha -N=\frac{m{v}^2}{r},N=0$

For N to vanish, both $\theta$ and v must increase a little more, Therefore with friction present, the mass will fly off the sphere at a greater angle than in part (i).