91Ó°ÊÓ

According to the work energy theorem, the net change in a body's energy is equal to the work that is done on it.

Here we have, the mass of block is\(m = 0.05\,{\rm{kg}}\).

The radius of circle is \(R = 0.8\,{\rm{m}}\).

The normal force on the block at the bottom of the circle is \(n = 3.4\,{\rm{N}}\) and the normal force on the block at the top of the circle is \({n_B} = 0.68\,{\rm{N}}\)

Here we have other forces than gravity is acting on the block.

So, work-energy theorem is given by:

\({K_1} + {U_1} + {W_{other}} = {K_2} + {U_2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 1 \right)\)

Where kinetic energy is given by:

\(K = \frac{1}{2}m{v^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 2 \right)\)

And the gravitational potential energy is given by:

\({U_{grav}} = mgy\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 3 \right)\)

We also know that, in a uniform circular motion, the acceleration is directed towards the center of the circle and its magnitude is given by:

\({a_{rad}} = \frac{{{v^2}}}{R}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 4 \right)\)

Now, apply Newton’s second law to the block at the bottom of the circle along the radial direction.

Therefore, we get

\(\begin{aligned}{}\sum {{F_{rad}} = n - mg = m{a_{rad}}} \\ \Rightarrow {a_{rad}} = \frac{n}{m} - g\end{aligned}\)

Put value of \({a_{rad}}\) in equation (4),

\(\begin{aligned}{}\frac{{{v^2}}}{R} &= \frac{n}{m} - g\\ \Rightarrow v &= \sqrt {\frac{{nR}}{m} - gR} \\ \Rightarrow v &= \sqrt {\frac{{3.4 \times 0.8}}{{0.05}} - 9.8 \times 0.8} &= 6.82\,{\rm{m/s}}\end{aligned}\)

Let\(y = 0\)at the bottom of the circle.

Now for point A at the bottom of circle

\( \Rightarrow {y_A} = 0,\;\;\;\;{v_A} = 6.82\,{\rm{m/s}}\)

For point B at the top of the circle

\({y_B} = 2R = 1.6\,{\rm{m}},\;\;\;\;{v_B} = ??{\rm{m/s}}\)

Here block start from zero gravitational potential energy level.

Therefore, \({U_1} = 0\)

We know that, in the uniform circular motion, the normal force is always perpendicular to the direction of motion,

\( \Rightarrow {W_{other}} = 0\)

Now, put value of \(m\;{\rm{and }}{{\rm{v}}_1}\) into equation (2),

\( \Rightarrow {K_1} = \frac{1}{2} \times 0.05 \times {\left( {6.82} \right)^2} = 1.16\,{\rm{J}}\)

Now, put value of \(m\;{\rm{and }}{{\rm{v}}_2}\) into equation (2),

\( \Rightarrow {K_2} = \frac{1}{2} \times 0.04 \times {\left( {{v_2}} \right)^2} = 0.025{v_2}^2\)

Now, put value of \(m\;{\rm{and }}{{\rm{y}}_2}\) into equation (3),

\( \Rightarrow {U_2} = 0.05 \times 9.8 \times 1.6 = 0.784\,{\rm{J}}\)

Now, put these values in equation (1),

\(\begin{aligned}{} \Rightarrow 1.16 + 0 + 0 = 0.025{v_2}^2 + 0.784\\ \Rightarrow {v_2}^2 &= \frac{{0.376}}{{0.025}}\\ \Rightarrow {v_2} &= 3.88\,{\rm{m/s}}\end{aligned}\)

Now, apply Newton’s second law to the block at the top of the circle along the radial direction.

Therefore, we get

\(\begin{aligned}{}\sum {{F_{rad}} = n + mg = m{a_{rad}}} \\ \Rightarrow n = m\left( {{a_{rad}} - g} \right)\end{aligned}\)

Put value of \({a_{rad}}\) in equation (4),

\(\begin{aligned}{}\frac{{{v_B}^2}}{R} &= \frac{{{n_B}}}{m} + g\\n &= m\left( {\frac{{{v^2}}}{R} - g} \right)\\ \Rightarrow n &= 0.05\left( {\frac{{{{\left( {3.88} \right)}^2}}}{{0.8}} - 9.8} \right)\\ \Rightarrow n &= 0.45\,{\rm{N}}\end{aligned}\)

Hence magnitude of the normal force that the track exerts on the block when it is at the top of its path is \(0.45\,{\rm{N}}\).