Question

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so that its pendulum motion takes 3.00 s. How far from the center of the rod should the pivot be located?

Answer 1

To find the distance from the center of the rod where the pivot should be located, calculate the period of the pendulum using the equation T = 2π√(l/g) and rearrange it to solve for l. Plug in the given values and solve to find l = 0.234 meters.

Step-by-step explanation:

To determine the distance from the center of the rod where the pivot should be located, we can use the equation for the period of a pendulum: T = 2π√(l/g), where T is the period, l is the length of the pendulum, and g is the acceleration due to gravity. Rearranging the equation, we have l = (T/2π)^2 * g. Plugging in the given values, we have l = (3.00/2π)^2 * 9.8 m/s^2, which gives us l = 0.234 meters.

Answer 2

0.087 m

Step-by-step explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

`T = 2\pi \sqrt{(I)/(mgd)}` .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

`I = (mL^(2))/(12)+ md^(2)`

Substituting the values in equation (1)

`3 = 2 \pi \sqrt{((mL^(2))/(12)+ md^(2))/(mgd)}`

`9=4\pi^(2)* \left ( ((L^(2))/(12)+d^(2))/(gd) \right )`

12d² -26.84 d + 2.25 = 0

`d=\frac{26.84\pm \sqrt{26.84^(2)-4* 12* 2.25}}{24}`

`d=(26.84\pm 24.75)/(24)`

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.