Question

A disc of radious 12cm and mass 50gm is suspended horizontally by a long wire
as its centre Its period T1 of angular SHM is measured to be 8.43 S. An irregu-
larly sheped object x is then hung from the same wire and its period T2 of angu-
lar SMM is found to be 4.76s. What is the moment of inertia of the object x about
its suspension axis ?​

Answer 1

Answer:

Spring exerts a force equal to Kx on the center of mass of the disc.

Friction acts on opposite direction to cause rolling without slipping.

Kx−f=Macm

fR=21MR2α

Solving gives f=3Kx

Hence restoring force acting on sphere=kx−f=32Kx

Hence ω=3M2K

Hence time period of oscillation=ω2π=2π2K3M

Answer 2

Correct question:

A disc of radius 12cm and a mass of 250g, is suspended horizontally by a long wire as its centre. Its period T1 of angular SHM is measured to be 8.43s. An irregularly shaped object x is then hung from the same wire and its period T2 is found to be 4.76s. What is the rotational inertia of the object x about its suspension axis?  

The rotational inertia of the object is $5.738\times 10^{-}^{4}kg.m^{2}$

Given:

Radius = 12cm

Mass = 50gm

Time = 8.43s, 4.76s.

To Find:

The rotational inertia of the object x about its suspension axis.

Solution:

Here, it is given that the radius of the wire is $12cm$.

Now, the MI of the disc about the rotational axis is $\frac{1}{2}MR^{2}$.

here, after putting the given values in the above formula.

we get,

$MI=\frac{1}{2}(0.25)(0.12)^{2}$

now, after solving the above equation.

we get,

$=\frac{144}{8}\times 10^{-}^{4}\\=1.8\times 10^{-}^{3} kg.m^{2}$

Here, we use the period of torsional oscillation, that is $T=2\pi \sqrt{\frac{I}{C} }$.

now, the torsional oscillation for the disc and for the object $x$ will be

$T_1=2\pi \sqrt{\frac{I_1}{C} } \\T_2=2\pi \sqrt{\frac{I_2}{C} }$

here, the torsional constant $c$ would be the same for both.

so, now we get,

$T_1^{2}=4\pi ^{2}\frac{I_1}{C}\\ T_2^{2}=4\pi ^{2}\frac{I_2}{C}$

now, $\frac{I_1}{2}=\frac{T_1^{2} }{T_2^{2} }$

$I_2=I_1(\frac{T_2}{T_1})^{2}$

here, after putting the values in the above formula.

we get,

$=(1.8\times 10^{-}^{3})(\frac{4.76}{8.43})^{2} \\=5.738\times 10^{-}^{4}kg.m^{2}$

Hence, the rotational inertia of the object is $5.738\times 10^{-}^{4}kg.m^{2}$.

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