Name: Power Transmitted by a Wave on a String
Uploaded: 2019-07-19
Duration: 7 min 4 s
Description: We know that a wave transfers energy from one point to another, but to find the power transmitted by a wave on a string, watch this video.

Question

A hexagonal pencil placed on an in lined plane with a slope $α$ at right angle $0$ the generatrix (i.e. the line of intersection of the plane and the horizontal surface remains at rest. The same pencil place parallel to the generatrix rolls down th lane.
Determine the angle $φ$ between the axis of the pencil and the generatrix of the inclined plane (Fig.) at which the pencil is still in equilibrium.

Accepted Answer

By hypothesis, the coefficient of sliding friction between the pencil and the inclined plane satisfies the condition $μ ⩾ \tan α$ . Indeed, the pencil put at right angles to the generatrix is in equilibrium, which means that $mgsin α = F_{f r}$ , where $m g$ is the force of gravity, and $F_{f r}$ is the force of friction. But $F_{fr} ⩽ μ m g \cos α$ . Consequently, $m g \sin α ⩽ μ m g \cos α$ , whence $μ ⩾ \tan α$ .

Thus, the pencil will not slide down the inclined plane for any value of the angle $φ$ .

The pencil may start rolling down at an angle $φ_{0}$ such that the vector of the force of gravity

"leaves" the region of contact between the pencil and the inclined plane (hatched region in Fig.). In order to find this angle, we project the centre of mass of the pencil (point $A$ ) on the inclined plane and mark the point of intersection of the vertical passing through the centre of mass and the inclined plane (point $B$ ). Obviously, points $A$ and $B$ will be at rest for different orientations of the pencil if its centre of mass remains stationary. In this case, $A B = 2 l \cos 30 ° \tan α$ , where $2 l$ is the side of a hexagonal cross section of the pencil, and $2 l \cos 30 °$ is the radius of the circle inscribed in the hexagonal cross section.

As long as point $B$ lies in the hatched region, the pencil will not roll down the plane.

Let us write the condition for the beginning of rolling down

$\frac{A D}{\cos φ_{0}} = A B,$ or $\frac{l}{\cos φ_{0}} = l \sqrt{3} \tan α$

whence

$φ_{0} = \arccos (\frac{1}{\sqrt{3} \tan α})$
Thus, if the angle $φ$ satisfies the condition $\arccos (\frac{1}{\sqrt{3} \tan α}) ⩽ φ ⩽ \frac{π}{2}$ the pencil remains in equilibrium. The expression for the angle $φ_{0}$ is meaningful provided that $\tan α > \frac{1}{\sqrt{3}}$ . The fact that the pencil put parallel to the generatrix rolls down indicates that $\tan α > \frac{1}{\sqrt{3}}$ (prove this).

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