## A thin rod of length 1.3 m and mass 88 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulu

Question

A thin rod of length 1.3 m and mass 88 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 2.39 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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3 years 2021-08-12T11:25:46+00:00 2 Answers 8 views 0

Explanation:

Given:

Length, L = 1.3 m

Mass, M = 88 g

= 0.088 kg

Angular velocity, w = 2.39 rad/s

Inertia of the rod = inertia of a cylinder = 1/12 M × L^2

Rotational inertia, I = Icm + I

= 1/12 M × L^2 + M × L/2

= 0.088 × (1/12 × 1.3^2 + 1.3/2)

= 0.088 × 0.79

= 0.05 kgm^2

Kinetic energy, E = 1/2 × l × w^2

= 1/2 × 0.05 × 2.39^2

= 0.142 J

B.

Using conservation of total mechanical energy,

Potential energy = kinetic energy

M × g × h = 0.142

h = 0.142/(9.8 × 0.088)

= 0.164 m

a) the rod’s kinetic energy at its lowest position is 0.39 J

b) 0.459 m far above that position the center of mass rises.

Explanation:

Given

L = 1.3m

m = 88g

The rotational inertia of the road about an axis passing through its fixed end is given as follow

I = Icm + m

a)

The kinetic energy of the rod at the lowest point is given by the following

K =

K is the kinetic energy of the rod at the lowest point

b)

From the conservation of the total mechanical energy of the rod

Ki + Ui = Kf + Uf

Ki = Uf – Ui + 0

mgh = K = 0.39

h = 0.39/mg = 0.39/0.088*9.8 = 0.459 m