Motion in a Plane

4.20

(a) The position is given by $\stackrel{?}{r}$ = 3. 0t ? − 2.0t 2 ? + 4.0 ?

The velocity v is given by $\stackrel{?}{v}$ = $\frac{d\stackrel{?}{r}}{dt}$ = $\frac{d}{dt}$ (3.0t ? − 2.0t 2 ? + 4.0 ?)

$\stackrel{?}{v}$ = 3.0 ? – 4.0t?

Acceleration $\stackrel{?}{a}$ = $\frac{d\stackrel{?}{v}}{dt}$ = $\frac{d}{dt}$ (3.0 ? – 4.0t?) = – 4.0?

(b) At t = 2.0s

$\stackrel{?}{v}$ = 3.0 ? – 4.0t?

The magnitude of velocity is given by $\left|\stackrel{?}{v}\right|$ = (3.02 +- 8.02)0.5 = 8.54 m/s

Direction, $\theta$ = ${\tan }^{-1}?\frac{v_y}{v_x}$ = ${\tan }^{-1}?\frac{8}{3}$ = 69.44 $°$

4.19

(a) The net acceleration of a particle in circular motion is always directed along the radius of the circle towards the centre (centripetal force), in the case of a circular uniform motion. Hence the statement is False.

(b) The centrifugal force direction is always along the tangent, hence the statement is True.

(c) In uniform circular motion, the direction of the acceleration vector points is always toward the centre of the circle. The average of these vectors over one cycle is a null vector. Hence the statement is True.

4.18 The radius of the loop, r = 1 km = 1000 m

Speed, v = 900 km/h = 250 m/s

Centripetal acceleration, AC = $\frac{v^2}{r}$ = 250 $*250$ / 1000 m/s² = 62.5 m/s² = $\frac{62.5}{9.817}$ g = 6.37g

4.17 Given, the length of the string, l = 80 cm, No. of revolution = 14, Time taken = 25 s

We know Frequency, v = $\frac{No.ofrevolution}{timetaken}$ = $\frac{14}{25}$ Hz

Angular frequency $\omega$ = 2?v = 2 $*$ $\frac{22}{7}$ $*\frac{14}{25}$ rad/s = 3.52 rad/s

Centripetal acceleration ac = ${\omega }^2$ r = ${3.52}^2*\frac{80}{100}$ m/s2 = 9.91 m/s2

The direction of acceleration is towards the centre.

4.16 We know for a projectile motion, horizontal range is given by

R = $\frac{u^2}{g}sin2\theta$ , where R is the horizontal range, u is the velocity and $\theta$ is the angle of projectile

So $\frac{u^2}{g}$ = 100/sin 2 $\theta$

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection $\theta$ = 45 $°$ , $\frac{u^2}{g}$ = 100 …….(1)

The ball will achieve max height when it is thrown vertically upward. For such motion, final velocity v = 0

From the equation $v^2$ - $u^2$ =2gH, where acceleration a = -g, we get

0 - $u^2$ = -2gH, H = $\frac{u^2}{2g}$ = $\frac{100}{2}$ = 50m

4.15 The speed of the ball, u = 40 m/s

Ceiling height, h = 25m

Since the ball is thrown, it will follow the path of a projectile. For projectile motion, the maximum height for a body projected at an angle $\theta$ is given by

h= $\frac{u^2{sin}^2\theta }{2g}$ , sin2 $\theta$ = (25 $*2*9.807)/(40*40)$ , $\theta =33.61°$

Horizontal range is given by,=

R = $\frac{u^2}{g}sin2\theta$ = ((40 $*40$ ) $*\sin ?2*33.61°$ )/9.81 = 150.38m

4.14

The velocity of the boat $V_b$ = 51 km/h, the velocity of the wind $V_w$ = 72 km/h

Flag is fluttering in N-E direction, so the $V_w$ direction is N-E.

When the ship begins to sail towards North, the flag will flutter along the direction of relative velocity of the wind w.r.t. the boat.

The angle between $V_w$ & - $V_b$ = 90 $°+45°$

tan $\beta$ = $\frac{V_{b}sin(90+45)}{V_w+V_{b}cos(90+45)}$ = $\frac{51\mathrm{s}\mathrm{i}\mathrm{n}?(90+45)}{72+51\mathrm{c}\mathrm{o}\mathrm{s}?(90+45)}$ = $\frac{36.062}{35.937}$

$\beta =$ tan-1( $\frac{36.062}{35.937}$ ) = 45.099 $°$

Angle with respect to East direction = 45.099 $°$ - 45 $°$ = 0.099 $°$

Hence the flag will flutter almost due East.

4.13 Swimming speed = 4 km/h, Width of the river = 1.0 km

Time required to cross the river = $\frac{Widthoftheriver}{Swimmingspeed}$ = $\frac{1}{4}$ h= 15 min

Speed of the river = 3 km/h

Distance covered by the river in 15 min = 3 $*\left(\frac{15}{60}\right)$ $*1000$ = 750 m

4.12 Let Vc = Velocity of the cyclist

Vr = Velocity of the rain

If V is the relative velocity, the woman must hold her umbrella in the direction of V.

V = Vr + (-Vc) = 30 + (-10) = 20 m/s

Tan $\theta$ = $\frac{Vc}{Vr}$ = $\frac{10}{30}$ , $\theta$ = 18 $°$