An object of mass 'm' is projected from origin in a vertical xy plane at an angle 45°$45^{\circ}$ with the x-axis with an initial velocity v₀$v_{0}$ The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [g is acceleration due to gravity]

Related Formula

Idisc = (1)/(2) M R²$$I_{\text{disc}} = \frac{1}{2} M R^2$$ [cite: 833]

I = Icm + M d² (Parallel Axis Theorem)$$I = I_{\text{cm}} + M d^2 \quad \text{(Parallel Axis Theorem)}$$ [cite: 848]

Core Logic

Let the original mass density per unit area be σ$\sigma$. Without any cavity, the moment of inertia is: [cite: 198, 833]

I₁ = (MR²)/(2)$$I_1 = \frac{MR^2}{2}$$ [cite: 833]

The mass of the removed small section scales directly with its cut-out area profile: [cite: 198, 834]

m = (M)/(π R²) × π ((R)/(3))² = (M)/(9)$$m = \frac{M}{\pi R^2} \times \pi \left(\frac{R}{3}\right)^2 = \frac{M}{9}$$ [cite: 198, 834, 844]

The center of mass of the removed disk sits at a distance d = R - (R)/(3) = (2R)/(3)$d = R - \frac{R}{3} = \frac{2R}{3}$ away from the primary center O$O$[cite: 198, 205, 848]. Calculating its partial moment of inertia about O$O$ via the parallel axis theorem: [cite: 199, 848]

I₂ = (m r²)/(2) + m d² = ((M)/(9)((R)/(3))²)/(2) + (M)/(9)((2R)/(3))²$$I_2 = \frac{m r^2}{2} + m d^2 = \frac{\frac{M}{9}\left(\frac{R}{3}\right)^2}{2} + \frac{M}{9}\left(\frac{2R}{3}\right)^2$$ [cite: 848]

I₂ = (MR²)/(162) + (4MR²)/(81) = (MR² + 8MR²)/(162) = (9MR²)/(162) = (MR²)/(18)$$I_2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2 + 8MR^2}{162} = \frac{9MR^2}{162} = \frac{MR^2}{18}$$ [cite: 848, 850]

Subtracting the removed component from the original configuration: [cite: 851]

I = I₁ - I₂ = (MR²)/(2) - (MR²)/(18) = (9MR² - MR²)/(18) = (8MR²)/(18) = (4)/(9)MR²$$I = I_1 - I_2 = \frac{MR^2}{2} - \frac{MR^2}{18} = \frac{9MR^2 - MR^2}{18} = \frac{8MR^2}{18} = \frac{4}{9}MR^2$$ [cite: 851]

Matching this with (4)/(x)MR²$\frac{4}{x}MR^2$, we find x = 9$x = 9$[cite: 200, 208, 851, 853].

Pattern Recognition

For uniform planar surfaces, mass always scales squarely with linear dimension changes (r arrow (r)/(3) m arrow (m)/(9)$r \rightarrow \frac{r}{3} \implies m \rightarrow \frac{m}{9}$)[cite: 198, 834, 844]. Always apply the parallel axis theorem to bring the component values to a unified reference point before executing addition or subtraction[cite: 848, 851].

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Core Logic

For a solid sphere, the moment of inertia about the center of mass is I = (2)/(5)mR²$I = \frac{2}{5}mR^2$. Since it rolls without slipping, the condition vcm = ω R$v_{\text{cm}} = \omega R$ holds.

Substituting I$I$ and ω$\omega$ into the ratio:

Pattern Recognition

The ratio of translational to rotational kinetic energy for any rolling body is given by mR²Icm$\frac{mR^2}{I_{\text{cm}}}$. For a solid sphere, this becomes (1)/(2/5) = (5)/(2)$\frac{1}{2/5} = \frac{5}{2}$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

Core Logic

For a solid sphere: Isolid = (2)/(5)MR² a₁ = (g θ)/(1 + 2/5) = (5)/(7)g θ$I_{\text{solid}} = \frac{2}{5}MR^2 \implies a_1 = \frac{g\sin\theta}{1 + 2/5} = \frac{5}{7}g\sin\theta$.

For a hollow sphere: Ihollow = (2)/(3)MR² a₂ = (g θ)/(1 + 2/3) = (3)/(5)g θ$I_{\text{hollow}} = \frac{2}{3}MR^2 \implies a_2 = \frac{g\sin\theta}{1 + 2/3} = \frac{3}{5}g\sin\theta$.

Comparing accelerations: a₁ > a₂$a_1 > a_2$

Since acceleration of the solid sphere is greater, it takes less time to descend the incline: t₁ < t₂$t_1 < t_2$

Pattern Recognition

Smaller moment of inertia mass distribution (more concentrated at the center) yields larger acceleration down an incline, meaning a quicker descent.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion

Related Formula

The linear acceleration a$a$ for pure rolling motion down an incline profile is:

Core Logic

For a uniform solid cylinder, the moment of inertia around its central axis is:

Step 1: Calculating Acceleration

Substitute θ = 45°$\theta = 45^{\circ}$ and the cylinder inertial factor into the formula :

Pattern Recognition

Solid cylinder rolls with acceleration matching (2)/(3) g θ$\frac{2}{3} g\sin\theta$. Since 45° = 1√(2)$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, this simplifies directly to √(2)g3$\frac{\sqrt{2}g}{3}$.

Chapter Mix

Class 11 Physics: System of Particles and Rotational Motion