Question

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Answer 1

Answer:

Question :

A road of mass M and length L is released. Find out value of angular acceleration α just after releasing.

Solution :

Given that,

Mass of rod = M

Length of rod = L

❖ Moment of inertia of a thin road about an axis passing through its one end and perpendicular to rod is given by, I = ML²/3

Force mg is acting on centre of mass. Distance of centre of mass from any of one end will be L/2 as rod is uniform

1) Torque acting on a body and angular acceleration produced in it are related as, τ = I α Hold on, our servers are swamped. Wait for your answer to fully load.

2) Torque acting on the rod due* to the force Mg is given by, τ = Mgr = Mg × L/2

By equating both equations;

➙ I α = Mg × L/2

➙ ML²/3 × α = Mg × L/2

➙ α = 3gL/2L²

➙ α = 3g/2L rad/s²

Explanation:

Answer 2

Hola mate

Here is your answer -

Mass of rod = M

Length of rod = L

Moment of inertia of a thin road about an axis passing through its one end and perpendicular to rod is given by, I = ML²/3

Force mg is acting on center of mass. Distance of center of mass from any of one end will be L/2 as rod is uniform

1) Torque acting on a body and angular acceleration produced in it are related as, τ = I α Hold on, our servers are swamped. Wait for your answer to fully load.

2) Torque acting on the rod due* to the force Mg is given by, τ = Mgr = Mg × L/2

By equating both equations;

I α = Mg × L/2

ML²/3 × α = Mg × L/2

α = 3gL/2L²

α = 3g/2L rad/s²