# A thin rod of length 0.83 m and mass 110 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendu

Question

A thin rod of length 0.83 m and mass 110 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 5.71 rad/s. Neglecting friction and air resistance, find (a) the rod’s kinetic energy at its lowest position and (b) how far above that position the center of mass rises.

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1 year 2021-08-18T18:04:02+00:00 2 Answers 10 views 0

Explanation:

length of rod, L = 0.83 m

mass of rod, M = 110 g = 0.11 kg

angular speed, ω = 5.71 rad/s

Moment of inertia of rod about the fixed point

I = Icm + Mh²

I = 1/12 M L² + M L²/4

I = 1/12 x 0.11 x 0.83 x 0.83 + 0.11 x 0.83 x 0.83 / 4

I = 0.0063 + 0.01895 = 0.0252 kgm²

a) kinetic energy of rod is given by

K = 1/2 Iω²

K = 0.5 x 0.0252 x 5.71 x 5.71 = 0.41 Joule

(b) The kinetic energy at the bottom  is equal to the potential energy at a height of the center of mass. Let the centre of mass rises upto height h.

mgh = K

0.11 x 9.8 x h = 0.41

h = 0.381 m

h = 38.1 cm

2. Explanation:

(a)  The given data is as follows.

Length of the rod, L = 0.83 m

Mass of the rod, m = 110 g = 0.11   (as 1 kg = 1000 g)

At the lowest point, angular speed of the rod ($$\omega$$) = 5.71 rad/s

First, we will calculate the rotational inertia of the rod about an axis passing through its fixed end is as follows.

I = $$I_{CM} + mh^{2}$$

= $$\frac{1}{12}mL^{2} + m(\frac{L}{2})$$

= $$\frac{1}{12} \times 0.11 \times (0.83)^{2} + 0.11 \times \frac{0.83}{2}$$

= 0.00631 + 0.415

= 0.42131 $$kg m^{2}$$

Therefore, kinetic energy of the rod at its lowest point will be calculated as follows.

K = $$\frac{1}{2}I \omega^{2}$$

= $$\frac{1}{2} \times 0.42131 kg m^{2} \times (5.71)^{2}$$

= 6.86 J

Hence, kinetic energy of the rod at its lowest point is 6.86 J.

(b)   According to the conservation of total mechanical energy of the rod, we have

$$K_{i} + U_{i} = K_{f} + U_{f}$$

$$K_{i} = U_{f} – U_{i}$$

or,      mgh = K = 6.86 J

Therefore,      h = $$\frac{6.86}{mg}$$

= $$\frac{0.63}{0.11 \times 9.8}$$

= 0.584 m

Hence, the center of mass rises 0.584 m far above that position.