Question

a stone of mass 2kg is tied to a string of length 0.5m if the breaking tension of the string is 900N then the maximum angular velocity the stone can have in uniform circulation motion is​

Answer 1

Given data :-

• A stone of mass 2kg is tied to a string of length 0.5m.

• The breaking tension of the string is 900N.

To find :-

• The maximum angular velocity the stone.

Solution : -

{According to given}

→ mass of the stone, m = 2 kg

→ Let, length of the string be the radius {because stone performing uniform circular motion }, r = 0.5 m

→ Breaking tension {force} of string, F = 900 N

{According to given}

The stone performing an uniform circular motion. so we know that centripetal force act on stone.

Here, we know

→ Angular velocity, w = v/r

→ v = w × r ....( 1 )

→ centripetal acceleration, a = v²/r

{from eq. ( 1 ) }

→ centripetal acceleration, a = (w r)²/r

→ centripetal acceleration, a = w² × r .....( 2 )

Now, we use formula of force

→ Force = mass × centripetal acceleration

→ F = m × w² × r

{from given and eq. ( 2 ) }

→ 900 = 2 × w² × 0.5

→ 900 = 1.0 × w² i.e.

→ w² = 900

→ w = √900

→ w = 30 rad/sec

Hence, the maximum angular velocity of the stone in uniform circulation motion is 30 rad/sec.

More info :-

• Centripetal force is the force required to provide centripetal acceleration to a particle to move it in a circular path.

• Angular velocity : The time rate of angular displacement of a particle performing circular motion os called angular velocity.