Question

Obtain an expression for potential energy of a particle performing simple harmonic motion. Hence evaluate the potenhtial energy (a) at mean position and (b) extreme position.
A horizontal disc is freely rotating about a transverse axit passing through its centre at the rate of 100 revolutions per minute. A 20 gram blob of wax falls on the disc and sticks to teh disc at a distance of 5 cm form its axis. Moment of inertia of the disc about its axit passing through its centre of mass is `2xx10^(-4)` kg `m^(2)`. Calculate the new frequency of rotation of the disc.

Answer 1

An osciliating particle possesses both types of energies potential as well sa kinetic. It possesse potential energy position.
Ptential energy : Condsider a particle of mass m executing S.H.M. Let x be its displacement from F acting upon the particle is proportinal to ans opposite to the displacement x, we have
`F = - kx`
Where, the constant k gives the force per unit displacement, the for can also be expressed in terms of potential energy U of the particle is proportionla to and opposite to the displacement x, we have
`F = -kx`
Where, the constant k gives the force per unit displacemenjt. The force can also be expressed in terms of potential energy U of the particle as :
`F= -(dU)/(dx)`
`"thus, "(dU)/(dx)=kx`
On integrating, we get
`U=1/2"kx"^2+C`
Where C is a constant. If we assume the ppotential energy zero in the equilibrium position, i.e., ,
If `U= 0` at `x=0,` then `C=0`.
Therefore, `U=1/2"kx"^2`
Potential energy, `U=1/2m omega^(2)x^(2)" "[because k = m omega^2]`
Hence for above it is clear that the potential energy of a particle performing S.H.M. is maximum and displacement is minimum i.r., x = 0 .
`"P.E."=1/2"m"omega^(2)x^(2)=0`
(b) Potential energy at extreme position : At extreme position, velocity of a particle executing S.H.M. is munumum and displacement is maximum, e.i., `x=-=a`
`"P.E."=1/2"m"omega^(2)x^(2)=0`

Given : `n_(1)="100 r.p.m."`
`=100/60=1*66"r.p.s.`
`m = 20 "gm"`
`h=r=5cm = 0*05m`
`I_(1)` = Moment of inertia of the disc about its axiws `=2xx10^(-4)"kgm"^(2)`.
`I_(2)` = Moment of inertia of the disc about same axis with To wax.
To find: `n_(2)= ? `
From formula,
`I_(2)=I_(1)+mr^(2)" [Formparallel axis theorem ]"`
`I_(1)omega_(1)=I_(2)omega_(2)`
` I_(1)(2pin_(1)) = (I_(1)+mr^(2))(2pin_(2))`
`I_(1)n_(1)=(I_(1)+mr^(2))n_(2)`
`n_(2)=(I_(1)n_(1))/(I_(1)+"mr"^(2))`
`n_(2)=(2xx10^(-4)xx166)/(2xx10^(-4)+(2xx10^(-2)xx0.05xx0.05))`
`=(2xx10^(-4)xx1*66)/(2xx10^(-4)+0.5xx10^(-4))`
`=(2xx10^(-4)xx1*66)/(2.5xx10^(-4))`
`1*328=1"r.p.s"`
`n_(2)=1" r.p.s"`