When a body slides down from rest along a smooth inclined plane making an angle of $$45^{\circ}$$ with the horizontal, it takes time $$T$$. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it is seen to take time $$p T$$, where $$p$$ is some number greater than 1. Calculate the coefficient of friction between the body and the rough plane.
Consider the diagram where a body slides down from along an inclined plane of inclination $$\theta\left(=45^{\circ}\right)$$.
On smooth inclined plane Acceleration of a body sliding down a smooth inclined plane
$$\begin{aligned} a & =g \sin \theta \\ \text{Here,}\quad\theta & =45^{\circ} \\ \therefore \quad a & =g \sin 45^{\circ}=\frac{g}{\sqrt{2}} \end{aligned}$$
Let the travelled distance be s.
Using equation of motion, $$s=u t+\frac{1}{2} a t^2$$, we get
$$\begin{aligned} & s=0 \cdot t+\frac{1}{2} \frac{g}{\sqrt{2}} T^2 \\ \text{or}\quad & s=\frac{g T^2}{2 \sqrt{2}}\quad \text{... (i)} \end{aligned}$$
On rough inclined plane Acceleration of the body $$a=g(\sin \theta-\mu \cos \theta)$$
$$\begin{aligned} & =g\left(\sin 45^{\circ}-\mu \cos 45^{\circ}\right) \\ & =\frac{g(1-\mu)}{\sqrt{2}} \quad\left(\text { As, } \sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right) \end{aligned}$$
Again using equation of motion, $$s=u t+\frac{1}{2} a t^2$$, we get
$$\begin{aligned} & s=0(p T)+\frac{1}{2} \frac{g(1-\mu)}{\sqrt{2}}(p T)^2 \\ \text{or}\quad & s=\frac{g(1-\mu) p^2 T^2}{2 \sqrt{2}} \quad \text{.... (ii)} \end{aligned}$$
From Eqs. (i) and (ii), we get
$$\frac{g T^2}{2 \sqrt{2}}=\frac{g(1-\mu) p^2 T^2}{2 \sqrt{2}}$$
or $$\quad (1-\mu) p^2=1$$
or $$\quad 1-\mu=\frac{1}{p^2}$$
or $$\quad \mu=\left(1-\frac{1}{p^2}\right)$$
Figure shows $$\left(v_x, t\right)$$, and $$\left(v_y, t\right)$$ diagrams for a body of unit mass. Find the force as a function of time.
Consider figure (a)
$$\begin{aligned} v_x & =2 t \text { for } 0 < t <1 \mathrm{~s} \\ & =2(2-t) \text { for } 1 < t < 2 \mathrm{~s} \\ & =0 \text { for } t>2 \mathrm{~s} \end{aligned}$$
From figure (b)
$$\begin{aligned} v_y & =t \text { for } 0 < t<1 \mathrm{~s} \\ & =1 \text { for } t>1 \mathrm{~s} \end{aligned}$$
$$\begin{aligned} \therefore \quad F_x & =m a_x=m \frac{d v_x}{d t} \quad \text{for } 0 < t < 1s\\ & =1 \times 2 \quad \text{for } 0 < t < 2s\\ & =1(-2) \quad \text{for } 2 < t \end{aligned}$$
$$\begin{aligned} F_y & =m a_y=m \frac{d v_y}{d t} \\ & =1 \times 1 \text { for } 0 < t <1 \mathrm{~s}=0 \text { for } 1 < t \\ \text{Hence,}\quad F & =F_x \hat{\mathbf{i}}+F_y \hat{\mathbf{j}} \quad \text{for } 0 < t < 1s\\ & =2 \hat{\mathbf{i}}+\hat{\mathbf{j}} \quad \text{for } 1 < t < 2s\\ & =-2 \hat{\mathbf{i}} \quad \text{for } t < 2s\\ & =0 \end{aligned}$$
A racing car travels on a track (without banking) $$A B C D E F A . A B C$$ is a circular arc of radius $$2 R.CD$$ and $$F A$$ are straight paths of length $$R$$ and $$D E F$$ is a circular arc of radius $$R=100 \mathrm{~m}$$. The coefficient of friction on the road is $$\mu=0.1$$. The maximum speed of the car is $$50 \mathrm{~ms}^{-1}$$. Find the minimum time for completing one round.
Balancing frictional force for centripetal force $$\frac{m v^2}{r}=f=\mu N=\mu m g$$ where, $$N$$ is normal reaction.
$$\therefore \quad v=\sqrt{\mu r g}\quad$$ (where, $$r$$ is radius of the circular track)
For path ABC $$\quad$$ Path length
$$\begin{aligned} & =\frac{3}{4}(2 \pi 2 R)=3 \pi R=3 \pi \times 100 \\ & =300 \pi \mathrm{m} \end{aligned}$$
$$\begin{aligned} v_1=\sqrt{\mu 2 R g} & =\sqrt{0.1 \times 2 \times 100 \times 10} \\ & =14.14 \mathrm{~m} / \mathrm{s} \\ \therefore \quad t_1 & =\frac{300 \pi}{14.14}=66.6 \mathrm{~s} \end{aligned}$$
$$\begin{aligned} \text { For path } D E F \quad \text { Path length } & =\frac{1}{4}(2 \pi R)=\frac{\pi \times 100}{2}=50 \pi \\ v_2 & =\sqrt{\mu R g}=\sqrt{0.1 \times 100 \times 10}=10 \mathrm{~m} / \mathrm{s} \\ t_2 & =\frac{50 \pi}{10}=5 \pi \mathrm{s}=15.7 \mathrm{~s} \end{aligned}$$
For paths, $$C D$$ and $$F A$$
$$\begin{aligned} \text { Path length } & =R+R=2 R=200 \mathrm{~m} \\ t_3 & =\frac{200}{50}=4.0 \mathrm{~s} \end{aligned}$$
$$\therefore$$ Total time for completing one round
$$t=t_1+t_2+t_3=66.6+15.7+4.0=86.3 \mathrm{~s}$$
The displacement vector of a particle of mass $$m$$ is given by $$\mathbf{r}(t)=\hat{\mathbf{i}} A \cos \omega t+\hat{\mathbf{j}} B \sin \omega t$$.
(a) Show that the trajectory is an ellipse.
(b) Show that $$F=-m \omega^2 \mathbf{r}$$.
(a) Displacement vector of the particle of mass $$m$$ is given by
$$\mathbf{r}(t)=\hat{\mathbf{i}} A \cos \omega t+\hat{\mathbf{j}} B \sin \omega t$$
$$\therefore \quad$$ Displacement along $$x$$-axis is,
$$\begin{aligned} x & =A \cos \omega t \\ \text{or}\quad\frac{x}{A} & =\cos \omega t \quad \text{... (i)} \end{aligned}$$
Displacement along $$y$$-axis is,
and $$\quad y=B \sin \omega t$$
or $$\quad \frac{y}{B}=\sin \omega t$$
Squaring and then adding Eqs. (i) and (ii), we get
$$\frac{x^2}{A^2}+\frac{y^2}{B^2}=\cos ^2 \omega t+\sin ^2 \omega t=1$$
This is an equation of ellipse.
Therefore, trajectory of the particle is an ellipse.
$$\begin{aligned} &\text { (b) Velocity of the particle }\\ &\begin{aligned} \mathbf{v} & =\frac{d \mathbf{r}}{d t}=\hat{\mathbf{i}} \frac{d}{d t}(A \cos \omega t)+\hat{\mathbf{j}} \frac{d}{d t}(B \sin \omega t) \\ & =\hat{\mathbf{i}}[A(-\sin \omega t) \cdot \omega]+\hat{\mathbf{j}}[B(\cos \omega t) \cdot \omega] \\ & =-\hat{\mathbf{i}} A \omega \sin \omega t+\hat{\mathbf{j}} B \omega \cos \omega t \end{aligned} \end{aligned}$$
$$\text { Acceleration of the particle }(\mathbf{a})=\frac{d \mathbf{v}}{d t}$$
$$\begin{aligned} \text{or}\quad\mathbf{a} & =-\hat{\mathbf{i}} A \omega \frac{d}{d}(\sin \omega t)+\hat{\mathbf{j}} B \omega \frac{d}{d t}(\cos \omega t) \\ & =-\hat{\mathbf{i}} A \omega[\cos \omega t] \cdot \omega+\hat{\mathbf{j}} B \omega[-\sin \omega t] \cdot \omega \\ & =-\hat{\mathbf{i}} A \omega^2 \cos \omega t-\hat{\mathbf{j}} B \omega^2 \sin \omega t \\ & =-\omega^2[\hat{\mathbf{i}} A \cos \omega t+\hat{\mathbf{j}} B \sin \omega t] \\ & =-\omega^2 \mathbf{r} \end{aligned}$$
$$\begin{aligned} &\therefore \text { Force acting on the particle, }\\ &F=m \mathbf{a}=-\mathbf{m} \omega^2 \mathbf{r}, \end{aligned}$$
Hence proved.
A cricket bowler releases the ball in two different ways
(a) giving it only horizontal velocity, and
(b) giving it horizontal velocity and a small downward velocity.
The speed $$v_s$$ at the time of release is the same. Both are released at a height $$H$$ from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.
(a) When ball is given only horizontal velocity Horizontal velocity at the time of release $$\left(u_x\right)=v_s$$
During projectile motion, horizontal velocity remains unchanged,
$$\begin{aligned} \text{Therefore, } \quad & v_x=u_x=v_s \\ \text{In vertical direction,} \quad & v_y^2=u_y^2+2 g H \\ v_y=\sqrt{2 g H} \quad \left(\because u_y=0\right) \end{aligned}$$
$$\begin{aligned} &\therefore \text { Resultant speed of the ball at bottom, }\\ &\begin{aligned} v & =\sqrt{v_x^2+v_y^2} \\ & =\sqrt{v_s^2+2 g H} \quad \text{... (i)} \end{aligned} \end{aligned}$$
$$\text { (b) When ball is given horizontal velocity and a small downward velocity }$$
Let the ball be given a small downward velocity $$u$$.
$$\begin{aligned} \text{In horizontal direction} \quad & v_x^{\prime}=u_x=v_s \\ \text{In vertical direction } \quad & v_y^{\prime 2}=u^2+2 g H \\ \text{or} \quad & v_y^{\prime}=\sqrt{u^2+2 g H} \end{aligned}$$
$$\therefore$$ Resultant speed of the ball at the bottom
$$v^{\prime}=\sqrt{v_x^{\prime 2}+v_y^{\prime 2}}=\sqrt{v_s^2+u^2+2 g H} \quad \text{... (ii)}$$
From Eqs. (i) and (ii), we get $$\quad v^{\prime}>v$$