A Block of mass 1kg is placed on a horizontal table rotating at a constant angular velocity w rad/sec at a distance 1m from the axis by connecting it to a string of length 1m whose other end is fixed at the center of the table.The value(s) of w for which tension in string is zero is/are:(the coefficient friction between block and surface is 0.4)
Answers
Answer:A mass m= 1 kg is revolved around a horizontal circle (thus no effect of gravity)
along a radius of r=1m
The mass experiences a centripetal force:
F
e
=mrω
2
[ω = Angular frequency]
This centripetal force is provided by the tension in string.
⇒T=F
e
=mrω
2
as given T=⊥6π
2
N, we get:
⊥6π
2
=(1)(⊥)ω
2
⇒ω
2
=⊥6π
2
⇒ω=4π
Thus, frequency of rotation,
ν=
2π
ω
=2 (Ans)
Answer:
Frequency of rotation is given by 2
Explanation:
Given:
A mass m= 1 kg is revolved around a horizontal circle (thus no effect of gravity) along a radius of r=1m
To find: frequency of rotation,
Solution:
The mass experiences a centripetal force:
Fe=mrω2
[ω = Angular frequency]
This centripetal force is provided by the tension in string.
⇒T=Fe=mrω2
as given T=⊥6π2N,
we get:
⊥6π2=(1)(⊥)ω2
⇒ω2=⊥6π2
⇒ω=4π
Thus, frequency of rotation,
ν= 2πω=2 (Ans)