A horizontal disc rotating freely about a vertical axis through its center makes \[90\] revolutions per minute. A small piece of wax of mass \[m\] falls vertically on the disc and sticks to it at a distance \[\;r\;\] from the axis. If the number of revolutions per minute reduces to\[60\;\], then the moment of inertia of the disc is?
Answer
563.4k+ views
Hint: The quantity that expresses a body’s tendency to resist angular acceleration is called the moment of inertia. It is the sum of mass times the square of the perpendicular distance to the axis of rotation. It can also be defined as the amount of torque needed for a specific angular acceleration around the rotational axis. Moment of inertia’s SI unit is given as \[kg/{m^2}\].
Complete step by step solution:
Given that the horizontal axis is rotating freely. Let the moment of inertia of the disc initially be \[I\]. The disc is rotating in the vertical axis through the center making \[90\]revolutions per minute.
Now a piece of mask falls on the disc vertically. The mass of the wax is \[m\]. This wax sticks to the disk at a distance of \[\;r\;\] from the axis.
Therefore the net moment of inertia is \[I + m{r^2}\]
After wax fell on the disc the no of rotations of the disc per minute reduces to \[60\;\]
There is no external torque acting on the system. Therefore the angular momentum remains constant.
The angular momentum formula is given as,
\[L = I\omega \]
Here \[L\] is the angular momentum.
Equating both angular momenta of the disc and disc with wax
\[I\omega \]=\[\omega (I + m{r^2})\]
\[1.5I = I + m{r^2}\]
\[1.5I - I = m{r^2}\]
\[0.5I = m{r^2}\]
\[I = \dfrac{1}{{0.5}}m{r^2}\]
\[I = 2m{r^2}\]
Therefore the amount of inertia of the disc is \[2m{r^2}\]
Note:
The role of the moment of inertia in rotational motion is the same as the role of mass in linear motion. The moment of inertia depends upon the density of the material, shape, and size of the material and the axis of rotation. The rotating bodies can be categorized as discrete, that is a system of particles and continuous, which refers to the rigid body.
Complete step by step solution:
Given that the horizontal axis is rotating freely. Let the moment of inertia of the disc initially be \[I\]. The disc is rotating in the vertical axis through the center making \[90\]revolutions per minute.
Now a piece of mask falls on the disc vertically. The mass of the wax is \[m\]. This wax sticks to the disk at a distance of \[\;r\;\] from the axis.
Therefore the net moment of inertia is \[I + m{r^2}\]
After wax fell on the disc the no of rotations of the disc per minute reduces to \[60\;\]
There is no external torque acting on the system. Therefore the angular momentum remains constant.
The angular momentum formula is given as,
\[L = I\omega \]
Here \[L\] is the angular momentum.
Equating both angular momenta of the disc and disc with wax
\[I\omega \]=\[\omega (I + m{r^2})\]
\[1.5I = I + m{r^2}\]
\[1.5I - I = m{r^2}\]
\[0.5I = m{r^2}\]
\[I = \dfrac{1}{{0.5}}m{r^2}\]
\[I = 2m{r^2}\]
Therefore the amount of inertia of the disc is \[2m{r^2}\]
Note:
The role of the moment of inertia in rotational motion is the same as the role of mass in linear motion. The moment of inertia depends upon the density of the material, shape, and size of the material and the axis of rotation. The rotating bodies can be categorized as discrete, that is a system of particles and continuous, which refers to the rigid body.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

