In Problem 6, what are the magnitudes of:

1. The horizontal component
2. The vertical component of the netforce acting on the block at point Q?
3. At what height hshould the block be released from rest so that it is on the verge of losing contact with the track at the top of the loop? (On the verge of losing contact means that the normal force on the block from the track has just then become zero.)
4. Graph the magnitude of the normal force on the block at the top of the loop versus initial height h, for the range h=0to h=6R.

1. Mass of block,$M=0.032Kg$
2. Loop radius,$R=12cm=0.12m$
3. Height of point P,$h=5.0R$

Using energy conversion and the concept of changing energy as per the position of block and forces acting on it, find the results. According to the law of energy conservation, energy can neither be created, nor be destroyed.

Formulae are as follow:

1. Kinetic energy,$KE=\frac{1}{2}m{v}^2$
2. Potential energy,$PE=mgh$
3. $KE+PE=cons\tan t$
   
4. Centripetal force, $F=\frac{m{v}^2}{R}$
5. Force, localid="1663048855896" $F=mg$where, KE is kinetic energy, PEis potential energy, m is mass, v is velocity, g is an acceleration due to gravity, F is force, R is radius, and h is height.

About the horizontal component at point Q, it is known that the block has centripetal force, which is along the radius toward the centre. It is given by,

$F=\frac{m{V}^2}{R}$leftward,

So, $K{E}_p+P{E}_p=K{E}_Q+P{E}_Q$

At point P, there is no kinetic energy. Therefore,

$\begin{array}{rcl}0+mgh& =& \frac{1}{2}m{v}^2+mgR\\ As,h& =& 5R\\ mg\left(5R\right)& =& \frac{1}{2}m{v}^2+mgR\\ m{v}^2& =& 8mgR\end{array}$

Now, centripetal force is,

role="math" localid="1663047227853" $\begin{array}{rcl}F& =& \frac{m{V}^2}{R}\\ F& =& \frac{8mgR}{R}\\ F& =& 8mg\\ F& =& 8\left(0.032\right)\left(9.8\right)\\ F& =& 2.5N\\ & & \end{array}$

Hence, horizontal component of net force acting on the block at point Q is 2.5N

The vertical component acting on the block is only its weight in downward direction,

$$\begin{gathered} F=mg \\ F=\left(0.032\right)\left(9.8\right) \\ F=0.31N \end{gathered}$$

Hence, vertical component of net force acting on the block at point Q is 0.31N.

As the condition given in the problem, for the loss of contact between block and track, the gravitational force must be equal to centripetal force,

$\begin{array}{rcl}\frac{m{v}^2}{R}& =& mg\\ m{v_{top}}^2& =& mgR\end{array}$

Law of conservation of energy,

$\begin{array}{rcl}K{E}_P+P{E}_P& =& K{E}_{top}+P{E}_{top}\\ 0+mgh& =& \frac{1}{2}m{v}_{top}^2+mg{h}_{top}\\ gh& =& \frac{1}{2}\left(gR\right)+g\left(2R\right)\\ h& =& \frac{1}{2}\left(R\right)+\left(2R\right)\\ h& =& \frac{5}{2}R\\ h& =& \frac{5}{2}\left(0.12\right)\\ h& =& 0.30m\\ & & \end{array}$

Hence, the block should be released from rest at height h=0.30 m so that it is on the verge of losing contact with the track at the top of the loop.

At the top of the loop,

$F_N=\frac{m{v_t}^2}{R}-mg$

$v_t$is the velocity at the top.

From law of conservation of energy,

localid="1663048893886" $\begin{array}{rcl}mgh& =& \frac{1}{2}m{v}^2+mg2R\\ gh& =& \frac{1}{2}{v}^2+2gR\end{array}$

To barely make it to the top, centripetal force would be equal to the force of gravity,

localid="1663048906045" $\begin{array}{rcl}\frac{m{v}^2}{R}& =& mg\\ v^2& =& Rg\end{array}$

So,

localid="1663048919661" $gh=\frac{Rg}{2}+2Rg$

That implies,

localid="1663048932222" $h\ge 2.5R$

Then the normal force, as a function of h, can be written as,

localid="1663048947102" $F_N=\frac{2mgh}{R}-5mg$

Using the above equation, localid="1663048962952" $F_N$vs h can be plotted between h=R to h=6R as follows:

Hence, graph of normal force on the block at the top versus height h (for h to 6R) is plotted.