Figure 8-32 shows a thin rod, of length L = 2.4 m and negligible mass, that can

Question

Figure 8-32 shows a thin rod, of length L = 2.4 m and negligible mass, that can pivot about one end to rotate in a vertical circle. A heavy ball of mass m = 14 kg is attached to the other end. The rod is pulled aside to angle ?0 = 20 and released with initial velocity 0 = 0. As the ball descends to its lowest point, (a) how much work does the gravitational force do on it and (b) what is the change in the gravitational potential energy of the ball-Earth system? (c) If the gravitational potential energy is taken to be zero at the lowest point, what is its value just as the ball is released?

Explanation / Answer

A thin rod of length L = 2.4 m and negligible mass

A heavy ball of mass m = 14 kg is attached to the other end.

The rod is pulled aside to angle 0 = 20° and released with initial velocity = 0

difference between height of lowest point and and extreme point from where it is released, h.

h = l-lcos

(a). work done by the gravitational force on it :

gravitation force = mg = 14*9.81 = 137.34 N

displacement along the gravitation force = l-lcos = 2.4-2.4*cos20 = 0.145 m

so work done = F*h = 137.34*0.145 = 19.9143 Joule

(b). change in the gravitational potential energy of the ball-Earth system :

change in potential energy = mgh2 - mgh1

assume extreme point line is reference line.

therefore, h1 = 0

h2 = 0.145 m

hence, change in potential energy = 14*9.81*0.145 - 14*9.81*0 = 19.9143 J

(c).

the gravitational potential energy is taken zero at the lowest point,

then its value at just the ball is released = mgh = 14*9.81*0.145 = 19.9143 J

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.