91Ó°ÊÓ

The mass of the potato (\(m\)) = 0.30 kg.

The length of the string (\(l\)) = 2.5 m.

Let us take \(h = 0\) at the potato’s lowest point.

Then its initial height is \({h_1} = l = 2.5\)m.

We know the work-energy theorem given by,

\({K_1} + {U_1} + W = {K_2} + {U_2}\,\,\, \cdots \cdots \left( 1 \right)\) ,

where the kinetic energy is given by,

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

And the gravitational potential energy is given by,

\(U = mgh\,\,\, \cdots \cdots \left( 3 \right)\).

From chapter 5, we know that, in circular motion, the acceleration vector is directed towards the center of the circle and its magnitude is given by,

\(a = \frac{{{v^2}}}{R}\,\,\, \cdots \cdots \left( 4 \right)\)

(a)

Since the tension in the rope is always perpendicular to the direction of motion, we have,

\(W = 0\).

Here, the potato started motion from the rest.

Therefore, \({K_1} = 0\).

Now, substituting the values of \(m,\,\,{h_1}\) in \(\left( 3 \right)\), we get,

\(U = 0.3 \times 9.8 \times 2.5 = 7.35\,{\rm{J}}\)

Since the potato ends at the zero potential level, we get,

\({U_2} = 0\).

Substituting all these values of work and energy quantities in \(\left( 1 \right)\), we get,

\(\begin{aligned}{}0 + 7.35 = {K_2} + 0\\ \Rightarrow {K_2} = 7.35J\end{aligned}\)

Now, substituting the values of \(m,\,\,{K_2}\) in \(\left( 2 \right)\), we get,

\(\begin{aligned}{}7.35 = \frac{1}{2} \times 0.3 \times {v_2}^2\\ \Rightarrow {v_2} = \sqrt {\frac{{2 \times 7.35}}{{0.3}}} \\ \Rightarrow {v_2} = 7\,\end{aligned}\)

\(\therefore {v_2} = 7\,\)m/s.

Hence, the speed is 7 m/s.

(b)

Since the potato is in a circular motion, at any point in its path, its acceleration is given by equation \(\left( 4 \right)\).

Therefore, by applying Newton’s Second Law to the potato at this point and along the radial direction, we get,

\(\begin{aligned}{}\sum {F = T - mg = m\frac{{{v^2}}}{l}} \\ \Rightarrow T = m\left( {g + \frac{{{v^2}}}{l}} \right)\end{aligned}\)

Then putting the values of \(m,\,{v_2},\,l\) , we get,

\(\begin{aligned}{}T = 0.3 \times \left( {9.8 + \frac{{{{\left( 7 \right)}^2}}}{{2.5}}} \right)\\ = 0.3 \times \left( {29.4} \right)\\ = 8.82\end{aligned}\)

\(\therefore T = 8.82\)N

Hence, the tension is 8.82 N.