Question #8098

A rod of length 56.0 cm and mass 1.40 kg is suspended by two strings which are 42.0 cm long, one at each end of the rod.
The string on side B is cut. Find the magnitude of the initial acceleration of end B.
The string on side B is retied and now has only half the length of the string on side A.
Find the magnitude of the initial acceleration of the end B when the string is cut.

Expert's answer

A rod of length 56.0 cm and mass 1.40 kg is suspended by two strings which are 42.0 cm long, one at each end of the rod.

&

The string on side B is cut. Find the magnitude of the initial acceleration of end B.

&

The string on side B is retied and now has only half the length of the string on side A.

&

Find the magnitude of the initial acceleration of the end B when the string is cut.

When the string is cut rod rotates about other end. Let's take moment equation about that end:

mg*l/2 = Iα [I = ml^2/3]

mgl/2 = ml^2/3 * α

Angular acceleration:

& α = 3g/2l

Acceleration of end B:

a = lα = 3g/2 = 14.7 [g=9.8].

&

The string on side B is cut. Find the magnitude of the initial acceleration of end B.

&

The string on side B is retied and now has only half the length of the string on side A.

&

Find the magnitude of the initial acceleration of the end B when the string is cut.

When the string is cut rod rotates about other end. Let's take moment equation about that end:

mg*l/2 = Iα [I = ml^2/3]

mgl/2 = ml^2/3 * α

Angular acceleration:

& α = 3g/2l

Acceleration of end B:

a = lα = 3g/2 = 14.7 [g=9.8].

## Comments

## Leave a comment