Question 424: Centre of Mass and Rotational Motion

Question

The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is $$\frac{1}{x} \mathrm{MR}^{2}$$, where $$\mathrm{R}$$ is the radius and $$M$$ is the mass of the semicircular ring. The value of $$x$$ will be __________.

Accepted Answer

To solve this problem, we need to understand the concept of moment of inertia. Moment of inertia is a measure of an object's resistance to rotational motion. It depends on the mass distribution of the object and the axis of rotation.
For a continuous object like a semicircular ring, we can calculate the moment of inertia by integrating over the entire object. Here's how we can approach this problem:
1.
Divide the semicircular ring into small mass elements:
Imagine the semicircular ring divided into infinitesimally small mass elements, each with mass $$dm$$.
2.
Calculate the moment of inertia of each element:
The moment of inertia of each element about the axis passing through the center and perpendicular to the plane of the ring is given by $$dI = dmR^2$$, where R is the radius of the ring.
3.
Integrate to find the total moment of inertia:
To find the total moment of inertia, we need to integrate $$dI$$ over the entire ring. This means integrating from $$0$$ to $$\pi$$ (the angle spanned by the semicircle) with respect to the angle $$	heta$$.
4.
Relate $$dm$$ to the total mass:
Since the ring has a uniform mass distribution, we can express the mass of each element $$dm$$ as a fraction of the total mass $$M$$: $$dm = \frac{M}{ÃÂR} Rd	heta = \frac{M}{\pi} d	heta$$.
Now, let's perform the integration:
$$I = \int_{0}^{\pi} dI = \int_{0}^{\pi} dmR^2 = \int_{0}^{\pi} \frac{M}{\pi} d	heta R^2$$
$$I = \frac{MR^2}{\pi} \int_{0}^{\pi} d	heta = \frac{MR^2}{\pi} [	heta]_{0}^{\pi}$$
$$I = \frac{MR^2}{\pi} [\pi - 0] = MR^2$$
Therefore, the moment of inertia of the semicircular ring about the given axis is $$MR^2$$. Comparing this to the given formula, we find that  $$x = \boxed{1}$$.

Centre of Mass and Rotational Motion

The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is $\frac{1}{x} \mathrm{MR}^{2}, where \mathrm{R} is the radius and M is the mass of the semicircular ring. The value of x$ will be __________.

Integer / Numeric Answer Type