Q48P Question: A rectangular block, w... [FREE SOLUTION]

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Chapter 15: Q48P (page 436)

Question: A rectangular block, with face lengths a = 35 cmand b = 45 cm, is to be suspended on a thin horizontal rod running through a narrow hole in the block. The block is then to be set swinging about the rod like a pendulum, through small angles so that it is in SHM. Figure shows one possible position of the hole, at distance rfrom the block鈥檚 center, along a line connecting the center with a corner.
Plot the period of the pendulum versus distance ralong that line such that the minimum in the curve is apparent.
For what value of rdoes that minimum occur? There is actually a line of points around the block鈥檚 center for which the period of swinging has the same minimum value.
What shape does that line make?

Short Answer

Expert verified

Answer

The period verses distance r along that line such that the minimum in the curve is apparent is plotted.
The value of r for minimum value of T is 0.16 m,
The shape made by the line is a circle.

Step by step solution

Identification of given data

The face lengths of the rectangular block is $a = 35 cm 鈥� 鈥� or 鈥� 鈥� 35 脳 10^{- 2} m$ and $b = 45 cm 鈥� 鈥� or 鈥� 鈥� 45 脳 10^{- 2} m$

Understanding the concept

The moment of inertia about an axis of rotation is equal to the sum of the moment of inertial about a parallel axis passing through the center of mass and the product of mass and a square of perpendicular distance between two axes. The time period of the physical pendulum can be defined in terms of its moment of inertia, mass, gravitational acceleration, and height.

Use the concept of parallel axis theorem and expression of the period for the physical pendulum.

Formulae:

$I = I_{c o m} + m h^{2}$

鈥�(颈)

Here, I is a moment of inertia about any axis, I_com is a moment of inertia about a parallel axis passing through the center of mass, is mass, and is the perpendicular distance between the two axes.

$T = 2 蟺 \sqrt{\frac{I}{m g h}}$

鈥�(颈颈)

Here, T is the time period, g is the gravitational acceleration

(a) Plot the period of the pendulum versus distance r along that line such that the minimum in the curve is apparent

According to the parallel axis theorem,

$I = I_{c o m} + m h^{2}$

Here, the distance of the hole from the center of the rectangular block is

The axis of rotation passing through the center of the rectangular block and perpendicular to its plane is

$I_{c o m} = \frac{m (a^{2} + b^{2})}{12}$

The equation (i) becomes as,

$I = \frac{m (a^{2} + b^{2})}{12} + m r^{2}$

The period of oscillation of the physical pendulum is,

$T = 2 蟺 \sqrt{\frac{I}{m g h}} = 2 蟺 \sqrt{\frac{\frac{m (a^{2} + b^{2})}{12} + m r^{2}}{m g r}} = 2 蟺 \sqrt{\frac{\frac{(a^{2} + b^{2})}{12} + r^{2}}{g r}} = \frac{2 蟺}{\sqrt{g}} \sqrt{\frac{(a^{2} + b^{2})}{12 r} + r}$

We plot the graph T verses r for the given a and b as

(b) Determining the value of at which the minimum occurs

For the minimum value of , its first derivative is zero.

$\frac{d T}{d r} = \frac{2 蟺}{\sqrt{g}} \sqrt{\frac{(a^{2} + b^{2})}{12} - r^{2}} 0 = \frac{2 蟺}{\sqrt{g}} \sqrt{\frac{(a^{2} + b^{2})}{12} - r^{2}} 0 = \frac{2 蟺}{\sqrt{g}} \sqrt{\frac{(a^{2} + b^{2})}{12} - r^{2}} r = \sqrt{\frac{(a^{2} + b^{2})}{12}} r = \sqrt{\frac{[{(35 脳 10^{- 2} m)}^{2} + {(45 脳 10^{- 2} m)}^{2}]}{12}} = 0.16 m$

The value of r at which minimum occur is 0.16m .

(c) Determining the shape made by the line

The direction from the center is not important. Hence, the locus of the point is a circle around the center of radius
$\sqrt{\frac{(a^{2} + b^{2})}{12}}$