A 0.300-kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. a) What is the speed of the potato at the lowest point of its motion? b) What is the tension in the string at this point?

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.