NCERT Solutions for Class 11 Physics Chapter 7 System of particles and Rotational Motion

Ans : Geometric centre; No 
The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. For the given geometric shapes having a uniform mass density, the C.M. 
lies at their respective geometric centres. The centre of mass of a body need not necessarily lie within it. For example, the C,M, of bodies such as a ring, a hollow sphere, etc., lies outside the body. 

Ans : No change The child is running arbitrarily on a trolley moving with velocity v. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the boy's motion is purely internal. Internal forces produce no effect on the moti«, cf the bodies on which they act. Since no external force is involved in the boy—trolley system, the boy's motion Hill produce no change in the velocity of the centre of mass of the trolley. 

Ans : \(\begin{array}{l}{\text { Let } \vec{a} \text { be represented } \vec{O P} \text { and } \vec{b} \text { be represented }} \\ {\text { by } \vec{O Q} . \text { Let } \angle P O Q=\theta, \text { Fig. }}\end{array}\)

\(\begin{array}{l}{\text { Complete the ll gm OPRQ. Join } P Q \text { . }} \\ {\text { Draw } Q N \perp O P} \\ {\text { In } \Delta O Q N, \quad \sin \theta=\frac{Q N}{O Q}=\frac{Q N}{b}}\end{array}\)
\(\begin{array}{c}{Q N=b \sin \theta} \\ {\text { Now, by definition, }|\vec{a} \times \vec{b}|=a b \sin \theta=(O P)(Q N)}\end{array}\)
\(\begin{aligned} &=\frac{2(O P)(Q N)}{2}=2 \times \text { area of } \Delta O P Q \\ \therefore \quad \text { area of } \Delta O P Q &=\frac{1}{2}\left|\vec{a}^{\prime} \times \vec{b}\right| \end{aligned}\)
which was to be proved. 

Ans : A parallelepiped with origin O and sides a, b, and c shown is in the following figure. 

\(\begin{array}{l}{\text { Volume of the given parallelepiped = abc }} \\ {\vec{\mathrm{OC}}=\vec{a}} \\ {\vec{\mathrm{OB}}=\vec{b}} \\ {\vec{\mathrm{OC}}=\vec{c}}\end{array}\)
\(\begin{array}{l}{\text { Let } \hat{\mathbf{n}} \text { be a unit vector perpendicular to both } b \text { and } c . \text { Hence, } \hat{\mathbf{n}} \text { and } a \text { have the same }} \\ {\text { direction. }}\end{array}\)
\(\begin{array}{l}{\therefore \vec{b} \times \vec{c}=b c \sin \theta \hat{\mathbf{n}}} \\ {=b c \sin 90^{\circ} \hat{\mathbf{n}}} \\ {=b c \hat{n}} \\ {\vec{a} \cdot(\vec{b} \times \vec{c})} \\ {=a \cdot(b c \hat{\mathbf{n}})}\end{array}\)
\(\begin{aligned} & \vec{a} \cdot(\vec{b} \times \vec{c}) \\ &=a \cdot(b c \hat{\mathbf{n}}) \\ &=a b c \cos \theta \hat{\mathbf{n}} \\ &=a b c \cos 0^{\circ} \\ &=a b c \\ &=\text { Volume of the parallelepiped } \end{aligned}\) 

Ans : \(\begin{aligned} l_{x} &=y \rho_{z}-z \rho_{y} \\ l_{y} &=2 p_{x}-x p_{z} \\ l_{z} &=x p_{y}-y p_{x} \end{aligned}\)
\(\begin{array}{l}{\text { Linear momentum of the particle, } \vec{p}=p_{x} \hat{\mathbf{i}}+p_{y} \mathbf{j}+p_{z} \hat{\mathbf{k}}} \\ {\text { Position vector of the particle, } \vec{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}} \\ {\text { Angular momentum, } \vec{l}=\vec{r} \times \vec{p}}\end{array}\)
\(=(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \times\left(p_{\mathbf{r}} \hat{\mathbf{i}}+p_{y} \hat{\mathbf{j}}+p_{r} \hat{\mathbf{k}}\right)\)
\(\left.\begin{aligned} \text { Comparing the coefficients of } \hat{\mathbf{i}}, \text { and } \hat{\mathbf{k}}, & \text { we get: } \\ l_{\mathrm{x}} &=y p_{x}-z p_{\mathrm{y}} \\ l_{y} &=x p_{z}-z p_{\mathrm{x}} \\ l_{z} &=x p_{y}-y p_{x} \end{aligned}\right\}\)
The particle moves in the x-v plane  Hence, the z-component of the position vector nd 
linear momentum vector becomes zero i.e., 
\(\left.\begin{array}{l}{l_{x}=0} \\ {l_{y}=0} \\ {l_{e}=x p_{y}-y p_{n}}\end{array}\right\}\)
Therefore, when the particle is confined to move in the x-V plane, the direction of angular momentum is along the z-direction.