Question

A 56-kg student runs at 6.0 m/s, grabs a hanging 10.0-m-longrope, and swings out over a lake (Fig. 6–50). He releases the rope when his velocity is zero. (a) What is the angle \({\bf{\theta }}\)when he releases the rope? (b) What is the tension in the rope just before he releases it? (c) What is the maximum tension in the rope during the swing?

FIGURE 6-50 Problem 92

Short answer

(a) The student releases the rope at an angle \(\theta = 35.31^\circ \). (b) Tension in the rope just before the student releases it is 448 N. (c) The maximum tension in the rope during the swing is 582 N.

Step-by-step solution

Conservation of mechanical energy

The law of conservation of mechanical energy states that whenever conservative forces act on a system, then total mechanical energy of the system remains constant,i.e., themechanical energy of the system at one point will be equal to the mechanical energy of the system at any other point during the motion. Thus,

\(\begin{aligned}K{E_1} + P{E_1} &= K{E_2} + P{E_2}\\\Delta KE + \Delta PE &= 0\end{aligned}\)

Given information

Mass of the student is\(m = 56\;{\rm{kg}}\).

The length of the rope is l = 10 m.

The initial speed of the student when he grabs the rope is\(u = 6\;{\rm{m/s}}\).

The final speed of the student when he releases the rope is\(v = 0\;{\rm{m }}{{\rm{s}}^{ - 1}}\).

Let the height of the rope above the ground whenthe student releases the rope is h, and the rope makes an angle \(\theta \) with the vertical. The various forces acting on the student and rope system is shown in the figure below:

(a) Determining the angle of the rope when student releases the rope.

The student grabs the rope when his velocity is u and releases the rope when his velocity is v.

So, the change in kinetic energy of the student is as follows:

\(\begin{aligned}\Delta KE &= \frac{1}{2}m\left( {{v^2} - {u^2}} \right)\\ &= - \frac{1}{2}m{u^2}\end{aligned}\)

Initially, the student is on the ground. Thus, the initial potential energy of the student is zero. At the point of release of rope, the student is at a height h.Thus, the final potential energy of the student is mgh. So, the change in potential energy of the student is as follows:

\(\begin{aligned}\Delta PE &= P{E_f} - P{E_i}\\ &= mgh - 0\end{aligned}\)\(\)

Therefore, according to the law of conservation of mechanical energy,

\(\begin{aligned}\Delta KE + \Delta PE &= 0\\\Delta KE &= - \Delta PE\\ - \frac{1}{2}m{u^2} &= - \left( {m{\rm{g}}h} \right)\\h &= \frac{{{u^2}}}{{2g}}\\ &= \frac{{{{\left( {6\;{\rm{m/s}}} \right)}^2}}}{{2\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}\\ &= 1.84\;{\rm{m}}\end{aligned}\)

From the figure, it is clear that

\(\begin{aligned}\cos \theta &= \frac{{l - h}}{l}\\ &= \frac{{\left( {10 - 1.84} \right)\;{\rm{m}}}}{{1.84\;{\rm{m}}}}\\ &= 0.816\\\theta &= {\cos ^{ - 1}}\left( {0.816} \right)\\ &= 35.31^\circ \end{aligned}\)

Thus, the student releases the rope at an angle \(\theta = 35.31^\circ \).

(b) Determining the tension in the rope just before releasing

The various forces acting on the rope at the point just before the student releases it is shown in the figure above. There will be no centripetal force acting on the rod at this point. The weight of the student (mg) acts vertically downwards; thus, the tension (T) in the rope acts in the upward direction.

It can be seen from the figure that the vertical component of tension balances the weight acting in the rope. Thus,

\(\begin{aligned}T &= mg\cos \theta \\T &= \left( {56\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {0.816} \right)\\ &= 447.8\;{\rm{N}}\\ &= 448\;{\rm{N}}\end{aligned}\)

Thus, the tension in the rope just before the student releases it is 448 N

(c) Determining the maximum tension in the rope

Let the tension in the rope is maximum at an angle \(\theta '\). At this angle, due to radial acceleration, centripetal force \({F_{\rm{C}}}\)must also be acting. Thus, according to Newton’s second law,

\(\begin{aligned}{F_{\rm{C}}} &= T - mg\cos \theta \\T &= {F_{\rm{C}}} + mg\cos \theta \\ &= \frac{{m{v^2}}}{l} + mg\cos \theta \end{aligned}\)

Here, v is the velocity of the student at the point of maximum tension.

It is clear from the above expression that tension in the rope will be maximum when the value of \(\cos \theta \) is maximum. However, the value of \(\cos \theta \)is maximum at \(0^\circ \). Thus, the tension in the rope is maximum when the student grabs the rope at the initial point.

Thus, maximum tension in the rope at \(0^\circ \)is as follows:

\(\begin{aligned}T &= \frac{{m{v^2}}}{l} + mg\cos 0^\circ \\ &= \frac{{\left( {56\;{\rm{kg}}} \right)\left( {6\;{\rm{m/s}}} \right)}}{{10.0\;{\rm{m}}}} + \left( {56\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( 1 \right)\\ &= 582.4\;{\rm{N}}\end{aligned}\)

Thus, the maximum tension in the rope during the swing is 582 N.