Explanation
The three rods form H, So let $$ H = I_1(AB) + I_2(EF) + I_3(CD)$$
Here we will find about left vertical axis i.e. $$I_1$$ AB
Moment of inertia of rod AB about the axis is $$0$$
Moment of inertia of rod CD(parallel axis theorem) about the axis is $$M{{L}^{2}}$$
Moment of inertia of rod EF (perpendicualr axis theorem) about the axis is $$\dfrac{1}{3}M{{L}^{2}}$$
So, Moment of inertia of the rod about the axis is
$$ =0+M{{L}^{2}}+\dfrac{1}{3}M{{L}^{2}} $$
$$ =\dfrac{4}{3}M{{L}^{2}} $$
Linear mass density of a rod AB ( of length 10 m) varies with distance x from its end A as $$\lambda = {\lambda _o}{x^3}$$$$\left( {{\lambda _o}\;{\text{is}}\;{\text{positive}}\;{\text{constant}}} \right)$$. Distance of the center of mass of the rod, from end B is
Find the coordination of center of mass of a uniform semicircle closed wire frame with respect to the origin which is at its center.The radius of the circular portion is R.
Correct answer is option (C).
Hint: Calculate the moment of inertia of each rod about a corner point. Then calculate the radius of gyration.
Step 1: Calculate moment of inertia of rods placed along $$AB$$ and $$AC$$.
Let us first have a look at the arrangement of the three rods each of length $$l$$ to form an equilateral triangle as shown below:
First, we calculate the moment of inertia of each rod. Since, all the rods are of equal length, so, the moment of inertia (MOI) of rod placed along $$AB$$ will be equal to that placed along $$AC$$.
By formula, MOI about an endpoint is given as $$\dfrac{m\times{l^2}}{3}$$ Therefore,
$$I_{AB}=I_{AC}=\dfrac{m{l^2}}{3}$$
Step 2: Calculate moment of inertia of rod placed along $$BC$$.
According to parallel axis theorem, the moment of inertia about an axis is equal to the sum of the moment of inertia about the parallel axis passing through the centre of mass of the body and the product of its mass and the square of the perpendicular distance ($$r$$) between the two parallel axes.
Thus, by using parallel axis theorem, we calculate the MOI of rod placed along $$BC$$
Therefore, $$I_{BC}=\dfrac{ml^2}{12}+mr^2$$
Now, from the figure of the triangle shown above, by applying Pythagoras theorem, we get,
$$r^2=l^2−\dfrac{l^2}{4}$$
$$\Rightarrow r^2=\dfrac{3l^2}{4}$$
Hence,
$$I_{BC}=\dfrac{ml^2}{12}+m\dfrac{3l^2}{4}$$
$$\Rightarrow I_{BC}=\dfrac{10ml^2}{12}$$
Step 3: Calculate the total moment of inertia of the three rods.
The total MOI will be:
$$I_t=I_{AB}+I_{AC}+I_{BC}$$
$$I_t=\dfrac{2ml^2}{3}+\dfrac{10ml^2}{12}$$
$$\Rightarrow I_t=\dfrac{3ml^2}{2}$$
Step 4: Calculate the radius of gyration.
The total mass of the system if $$3m$$
The radius of gyration can be written as:
$$Mk^2=I_t$$ (where, $$M$$ is the total mass of the system)
$$\Rightarrow3mk^2=\dfrac{3ml^2}{2}$$
$$\Rightarrow k^2=\dfrac{l^2}{2}$$
On solving, we get,
$$k=\dfrac{l}{\sqrt{2}}$$
Hence, the radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is $$\dfrac{l}{\sqrt{2}}$$
The correct answer is option (C).
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