Two iron solid discs of negligible thickness have radii R₁$\mathbf{R}_1$ and R₂$\mathbf{R}_2$ and moment of inertia I₁$\mathrm{I}_1$ and I₂$\mathrm{I}_2$ , respectively. For R₂ = 2R₁$\mathrm{R}_2 = 2\mathrm{R}_1$ , the ratio of I₁$\mathrm{I}_1$ and I₂$\mathrm{I}_2$ would be 1 / x$1 / \mathrm{x}$ , where x =$\mathrm{x} =$

Related Formula

• Moment of inertia of a circular ring of mass M$M$ and radius R$R$ about a diametrical axis in its plane:

• Parallel Axis Theorem:

where d$d$ is the perpendicular distance between the center of mass axis and the parallel axis.

Core Logic

The axis of rotation is tangential and lies in the plane of the ring. Thus, the distance from the center of mass axis (which is also diametrical and in-plane) is equal to the radius R$R$.

Using the parallel axis theorem:

Step 1: Express in terms of diameter r$r$

The question specifies the diameter of the ring is r$r$. Therefore, the radius R$R$ is:

Substitute R = (r)/(2)$R = \frac{r}{2}$ into the moment of inertia formula:

Thus, the moment of inertia is (3)/(8) M r²$\frac{3}{8} M r^2$.

Pattern Recognition

Sees: Circular ring tangential axis in-plane. Trap: Directly using radius r$r$ instead of diameter r$r$. Read variables carefully: often r$r$ is radius, but here it is explicitly given as diameter! Shortcut: Standard in-plane tangent is (3)/(2)MR²$\frac{3}{2}MR^2$. Substitute R = (r)/(2)$R = \frac{r}{2}$ to get (3)/(8)Mr²$\frac{3}{8}Mr^2$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Core Logic

Let the side length of the square lamina be l = 10~cm$l = 10\mathrm{~cm}$. The lamina is pivoted at point O(0,0)$O(0,0)$ and remains stationary under rotational equilibrium. Therefore, the net torque about O$O$ must be zero.

Evaluating torque contributions about point O$O$:

• Forces acting directly at pivot O$O$ produce zero torque.
• Forces whose lines of action pass through O$O$ produce zero torque.
• The 10~N$10\mathrm{~N}$ force perpendicular to side OA$OA$ produces torque:

• The unknown force F$F$ acting perpendicular to side OC$OC$ produces torque:

Setting Σ τO = 0$\sum \tau_O = 0$:

Step 1: Final Conclusion

The magnitude of the force F$F$ is 10~N$10\mathrm{~N}$.

Pattern Recognition

In pivoted laminas, focus on the pivot and disregard any force vector whose line of action passes through the pivot. For symmetric placements, equate Clockwise torque = Counter-Clockwise torque directly.

Chapter Mix

Class 11 Physics: Rotational Motion

Related Formula

WF = F · s$W_F = F \cdot s$

Core Logic

The work done by the constant pulling force F = 20~N$F = 20\mathrm{~N}$ through distance s = 1~m$s = 1\mathrm{~m}$ is completely converted into the rotational kinetic energy of the wheel.

Work done by force:

Since the spokes are of negligible mass, all mass M = 10~kg$M = 10\mathrm{~kg}$ is distributed on the outer rim of radius R = 10~cm = 0.1~m$R = 10\mathrm{~cm} = 0.1\mathrm{~m}$. The wheel acts as a thin ring:

Using the work-energy theorem:

Step 1: Final Conclusion

The angular velocity of the wheel after the cord is unwound by 1~m$1\mathrm{~m}$ is 20~rad/s$20\mathrm{~rad/s}$.

Pattern Recognition

Work done in unwinding a string is F · s$F \cdot s$. Under pure rotation with zero friction, this is exactly (1)/(2) I ω²$\frac{1}{2} I \omega^2$. Always identify the mass distribution (here, rim with massless spokes acts as a thin cylinder/ring I = MR²$I = MR^2$).

Chapter Mix

Class 11 Physics: Rotational Motion Class 11 Physics: Work, Energy and Power

Related Formula

Core Logic

Initially, the moment of inertia is:

When cut into two equal parts, each smaller rod has:

• Mass, m = (M)/(2)$m = \frac{M}{2}$
• Length, l = (L)/(2)$l = \frac{L}{2}$

When joined symmetrically as a cross, the target axis passes through their joint intersection perpendicular to their plane. For each rod, this axis passes through its individual center of mass and is perpendicular to its length.

Thus, the total moment of inertia of the cross is the sum of the moments of inertia of the two rods:

Substituting m = (M)/(2)$m = \frac{M}{2}$ and l = (L)/(2)$l = \frac{L}{2}$:

Comparing with α$\alpha$:

Related Formula

For a continuous system modeled as discrete point masses located at their respective centers of mass:

Core Logic

Let the uniform linear mass density of the rod be λ$\lambda$.

• Total length is 5L$5L$.
• One segment of length 2L$2L$ lies on the x-axis. Its mass is 2m = λ(2L)$2m = \lambda(2L)$ and its center of mass is at (L, 0)$(L, 0)$.
• The remaining segment of length 3L$3L$ lies on the y-axis. Its mass is 3m = λ(3L)$3m = \lambda(3L)$ and its center of mass is at (0, 1.5L)$(0, 1.5L)$.

Step 1: Calculate COM Coordinates

Find the coordinates of the system's center of mass:

Given L = 10 ~cm$L = 10 \mathrm{~cm}$:

Step 2: Vector Form

Expressing in vector notation:

Pattern Recognition

Sees: L-shaped rod formed by bending a total length Ltotal$L_{\text{total}}$. Shortcut: Treat each arm as a point mass at its geometric midpoint. For segments of ratio 2:3$2:3$, the COM divides the distance between their midpoints in the inverse ratio 3:2$3:2$ closer to the heavier segment on the y-axis.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion