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PHXI04:MOTION IN A PLANE

361883 Two racing cars of masses $m_{1}$ and $m_{2}$ are moving in circles of radii $r_{1}$ and $r_{2}$ respectively. Their speeds are such that each makes a complete circle in the same duration of time $t$ . The ratio of the angular speed of the first to the second car is

1

r_{1} : r_{2}

2

m_{1} : m_{2}

3

m_{1} r_{1} : m_{2} r_{2}

4

1 : 1

Explanation:

As time periods are equal therefore ratio of angular speeds will be $1 : 1. (ω = \frac{2 π}{T}) .$

PHXI04:MOTION IN A PLANE

361884 A pont $P$ moves in counterclockwise direction on a circular path as shown in figure. The movement of $P$ is such that it sweeps out a length $S = t^{3} + 8$ , where $S$ in metre and $t$ is in seconds. The radius of the path is $60 m$ . The acceleration of $P$ when $t = 4 s$ is near by:
supporting img

1

45.28 m / s^{2}

2

60 m / s^{2}

3

50.2 m / s^{2}

4

30 m / s^{2}

Explanation:

$S = t^{3} + 8$
$∴$ speed, $v = \frac{d s}{d t} = 3 t^{2} ∴ a = \frac{d v}{d t} = 6 t$
tangential acceleration $a_{t} = 6 \times 4 = 24 m / s^{2}$
Centripetal acceleration $a_{c} = \frac{v^{2}}{R} = \frac{{(3 t^{2})}^{2}}{R} = \frac{9 t^{4}}{R}$
$a_{c} = \frac{9 \times 16 \times 16}{60} = \frac{3 \times 16 \times 16}{20} a_{c} = 38.4 m / s^{2}$
$∴$ Net acceleration
$a = \sqrt{a_{c}^{2} + a_{t}^{2}} = \sqrt{(38.4)^{2} + (24)^{2}}$
$a = \sqrt{1474.56 + 576} = \sqrt{2050.56}$
$\Rightarrow a = 45.28 m / s^{2}$

PHXI04:MOTION IN A PLANE

361885 A particle moves in a circle of radius 4 $m$ clockwise at constant speed of $2 m s^{- 1} .$ If $\hat{x}$ and $\hat{y}$ are unit vectors along $x$ and $y$ axis, respectively, the acceleration of the particle at the instant half way between $P Q$ is given by
supporting img

1

- 4 (\hat{x} - \hat{y})

2

4 (\hat{x} + \hat{y})

3

\frac{- (\hat{x} + \hat{y})}{\sqrt{2}}

4

\frac{(\hat{x} - \hat{y})}{4}

Explanation:

Mid point acceleration,
$a = \frac{υ^{2}}{r} = \frac{4}{4} = 1 m s^{- 1}$ at $45^{\circ}$ with $x$ axis
$∴ {\vec{a}}_{x} = - a \cos 45^{\circ} \hat{x}$
$= - 1 \times \frac{1}{\sqrt{2}} \hat{x} = - \frac{1}{\sqrt{2}} \hat{x}$
${\vec{a}}_{y} = - a \sin 45^{\circ} \hat{y}$
$= - 1 \times \frac{1}{\sqrt{2}} \hat{y} = - \frac{1}{\sqrt{2}} \hat{y}$
Thus $\hat{a} = {\vec{a}}_{x} + {\vec{a}}_{y} = - \frac{1}{\sqrt{2}} (\hat{x} + \hat{y})$
supporting img

PHXI04:MOTION IN A PLANE

361886 A ball moves along an uneven horizontal road with fixed speed at all points on the road. The normal reaction of the road on the body at two different points $A$ and $B$ is
supporting img

1 Maximum at

A

2 Maximum at

B

3 Minimum at

B

4 Same at

A

and

B

Explanation:

Centripetal acceleration at $A$ and $B$ is
$\frac{m u^{2}}{r_{A}}$ and $\frac{m u^{2}}{r_{B}}$ respectively.
Clearly $r_{B} > r_{A}$ therefore centripetal acceleration at $A$ is more. Thus, reaction at $B$ is less than $A$ because here the normal reaction provides the centripetal force.

PHXI04:MOTION IN A PLANE

361883 Two racing cars of masses $m_{1}$ and $m_{2}$ are moving in circles of radii $r_{1}$ and $r_{2}$ respectively. Their speeds are such that each makes a complete circle in the same duration of time $t$ . The ratio of the angular speed of the first to the second car is

1

r_{1} : r_{2}

2

m_{1} : m_{2}

3

m_{1} r_{1} : m_{2} r_{2}

4

1 : 1

Explanation:

As time periods are equal therefore ratio of angular speeds will be $1 : 1. (ω = \frac{2 π}{T}) .$

PHXI04:MOTION IN A PLANE

361884 A pont $P$ moves in counterclockwise direction on a circular path as shown in figure. The movement of $P$ is such that it sweeps out a length $S = t^{3} + 8$ , where $S$ in metre and $t$ is in seconds. The radius of the path is $60 m$ . The acceleration of $P$ when $t = 4 s$ is near by:
supporting img

1

45.28 m / s^{2}

2

60 m / s^{2}

3

50.2 m / s^{2}

4

30 m / s^{2}

Explanation:

$S = t^{3} + 8$
$∴$ speed, $v = \frac{d s}{d t} = 3 t^{2} ∴ a = \frac{d v}{d t} = 6 t$
tangential acceleration $a_{t} = 6 \times 4 = 24 m / s^{2}$
Centripetal acceleration $a_{c} = \frac{v^{2}}{R} = \frac{{(3 t^{2})}^{2}}{R} = \frac{9 t^{4}}{R}$
$a_{c} = \frac{9 \times 16 \times 16}{60} = \frac{3 \times 16 \times 16}{20} a_{c} = 38.4 m / s^{2}$
$∴$ Net acceleration
$a = \sqrt{a_{c}^{2} + a_{t}^{2}} = \sqrt{(38.4)^{2} + (24)^{2}}$
$a = \sqrt{1474.56 + 576} = \sqrt{2050.56}$
$\Rightarrow a = 45.28 m / s^{2}$

PHXI04:MOTION IN A PLANE

361885 A particle moves in a circle of radius 4 $m$ clockwise at constant speed of $2 m s^{- 1} .$ If $\hat{x}$ and $\hat{y}$ are unit vectors along $x$ and $y$ axis, respectively, the acceleration of the particle at the instant half way between $P Q$ is given by
supporting img

1

- 4 (\hat{x} - \hat{y})

2

4 (\hat{x} + \hat{y})

3

\frac{- (\hat{x} + \hat{y})}{\sqrt{2}}

4

\frac{(\hat{x} - \hat{y})}{4}

Explanation:

Mid point acceleration,
$a = \frac{υ^{2}}{r} = \frac{4}{4} = 1 m s^{- 1}$ at $45^{\circ}$ with $x$ axis
$∴ {\vec{a}}_{x} = - a \cos 45^{\circ} \hat{x}$
$= - 1 \times \frac{1}{\sqrt{2}} \hat{x} = - \frac{1}{\sqrt{2}} \hat{x}$
${\vec{a}}_{y} = - a \sin 45^{\circ} \hat{y}$
$= - 1 \times \frac{1}{\sqrt{2}} \hat{y} = - \frac{1}{\sqrt{2}} \hat{y}$
Thus $\hat{a} = {\vec{a}}_{x} + {\vec{a}}_{y} = - \frac{1}{\sqrt{2}} (\hat{x} + \hat{y})$
supporting img

PHXI04:MOTION IN A PLANE

361886 A ball moves along an uneven horizontal road with fixed speed at all points on the road. The normal reaction of the road on the body at two different points $A$ and $B$ is
supporting img

1 Maximum at

A

2 Maximum at

B

3 Minimum at

B

4 Same at

A

and

B

Explanation:

Centripetal acceleration at $A$ and $B$ is
$\frac{m u^{2}}{r_{A}}$ and $\frac{m u^{2}}{r_{B}}$ respectively.
Clearly $r_{B} > r_{A}$ therefore centripetal acceleration at $A$ is more. Thus, reaction at $B$ is less than $A$ because here the normal reaction provides the centripetal force.

PHXI04:MOTION IN A PLANE

361883 Two racing cars of masses $m_{1}$ and $m_{2}$ are moving in circles of radii $r_{1}$ and $r_{2}$ respectively. Their speeds are such that each makes a complete circle in the same duration of time $t$ . The ratio of the angular speed of the first to the second car is

1

r_{1} : r_{2}

2

m_{1} : m_{2}

3

m_{1} r_{1} : m_{2} r_{2}

4

1 : 1

Explanation:

As time periods are equal therefore ratio of angular speeds will be $1 : 1. (ω = \frac{2 π}{T}) .$

PHXI04:MOTION IN A PLANE

361884 A pont $P$ moves in counterclockwise direction on a circular path as shown in figure. The movement of $P$ is such that it sweeps out a length $S = t^{3} + 8$ , where $S$ in metre and $t$ is in seconds. The radius of the path is $60 m$ . The acceleration of $P$ when $t = 4 s$ is near by:
supporting img

1

45.28 m / s^{2}

2

60 m / s^{2}

3

50.2 m / s^{2}

4

30 m / s^{2}

Explanation:

$S = t^{3} + 8$
$∴$ speed, $v = \frac{d s}{d t} = 3 t^{2} ∴ a = \frac{d v}{d t} = 6 t$
tangential acceleration $a_{t} = 6 \times 4 = 24 m / s^{2}$
Centripetal acceleration $a_{c} = \frac{v^{2}}{R} = \frac{{(3 t^{2})}^{2}}{R} = \frac{9 t^{4}}{R}$
$a_{c} = \frac{9 \times 16 \times 16}{60} = \frac{3 \times 16 \times 16}{20} a_{c} = 38.4 m / s^{2}$
$∴$ Net acceleration
$a = \sqrt{a_{c}^{2} + a_{t}^{2}} = \sqrt{(38.4)^{2} + (24)^{2}}$
$a = \sqrt{1474.56 + 576} = \sqrt{2050.56}$
$\Rightarrow a = 45.28 m / s^{2}$

PHXI04:MOTION IN A PLANE

361885 A particle moves in a circle of radius 4 $m$ clockwise at constant speed of $2 m s^{- 1} .$ If $\hat{x}$ and $\hat{y}$ are unit vectors along $x$ and $y$ axis, respectively, the acceleration of the particle at the instant half way between $P Q$ is given by
supporting img

1

- 4 (\hat{x} - \hat{y})

2

4 (\hat{x} + \hat{y})

3

\frac{- (\hat{x} + \hat{y})}{\sqrt{2}}

4

\frac{(\hat{x} - \hat{y})}{4}

Explanation:

Mid point acceleration,
$a = \frac{υ^{2}}{r} = \frac{4}{4} = 1 m s^{- 1}$ at $45^{\circ}$ with $x$ axis
$∴ {\vec{a}}_{x} = - a \cos 45^{\circ} \hat{x}$
$= - 1 \times \frac{1}{\sqrt{2}} \hat{x} = - \frac{1}{\sqrt{2}} \hat{x}$
${\vec{a}}_{y} = - a \sin 45^{\circ} \hat{y}$
$= - 1 \times \frac{1}{\sqrt{2}} \hat{y} = - \frac{1}{\sqrt{2}} \hat{y}$
Thus $\hat{a} = {\vec{a}}_{x} + {\vec{a}}_{y} = - \frac{1}{\sqrt{2}} (\hat{x} + \hat{y})$
supporting img

PHXI04:MOTION IN A PLANE

361886 A ball moves along an uneven horizontal road with fixed speed at all points on the road. The normal reaction of the road on the body at two different points $A$ and $B$ is
supporting img

1 Maximum at

A

2 Maximum at

B

3 Minimum at

B

4 Same at

A

and

B

Explanation:

Centripetal acceleration at $A$ and $B$ is
$\frac{m u^{2}}{r_{A}}$ and $\frac{m u^{2}}{r_{B}}$ respectively.
Clearly $r_{B} > r_{A}$ therefore centripetal acceleration at $A$ is more. Thus, reaction at $B$ is less than $A$ because here the normal reaction provides the centripetal force.

PHXI04:MOTION IN A PLANE

361883 Two racing cars of masses $m_{1}$ and $m_{2}$ are moving in circles of radii $r_{1}$ and $r_{2}$ respectively. Their speeds are such that each makes a complete circle in the same duration of time $t$ . The ratio of the angular speed of the first to the second car is

1

r_{1} : r_{2}

2

m_{1} : m_{2}

3

m_{1} r_{1} : m_{2} r_{2}

4

1 : 1

Explanation:

As time periods are equal therefore ratio of angular speeds will be $1 : 1. (ω = \frac{2 π}{T}) .$

PHXI04:MOTION IN A PLANE

361884 A pont $P$ moves in counterclockwise direction on a circular path as shown in figure. The movement of $P$ is such that it sweeps out a length $S = t^{3} + 8$ , where $S$ in metre and $t$ is in seconds. The radius of the path is $60 m$ . The acceleration of $P$ when $t = 4 s$ is near by:
supporting img

1

45.28 m / s^{2}

2

60 m / s^{2}

3

50.2 m / s^{2}

4

30 m / s^{2}

Explanation:

$S = t^{3} + 8$
$∴$ speed, $v = \frac{d s}{d t} = 3 t^{2} ∴ a = \frac{d v}{d t} = 6 t$
tangential acceleration $a_{t} = 6 \times 4 = 24 m / s^{2}$
Centripetal acceleration $a_{c} = \frac{v^{2}}{R} = \frac{{(3 t^{2})}^{2}}{R} = \frac{9 t^{4}}{R}$
$a_{c} = \frac{9 \times 16 \times 16}{60} = \frac{3 \times 16 \times 16}{20} a_{c} = 38.4 m / s^{2}$
$∴$ Net acceleration
$a = \sqrt{a_{c}^{2} + a_{t}^{2}} = \sqrt{(38.4)^{2} + (24)^{2}}$
$a = \sqrt{1474.56 + 576} = \sqrt{2050.56}$
$\Rightarrow a = 45.28 m / s^{2}$

PHXI04:MOTION IN A PLANE

361885 A particle moves in a circle of radius 4 $m$ clockwise at constant speed of $2 m s^{- 1} .$ If $\hat{x}$ and $\hat{y}$ are unit vectors along $x$ and $y$ axis, respectively, the acceleration of the particle at the instant half way between $P Q$ is given by
supporting img

1

- 4 (\hat{x} - \hat{y})

2

4 (\hat{x} + \hat{y})

3

\frac{- (\hat{x} + \hat{y})}{\sqrt{2}}

4

\frac{(\hat{x} - \hat{y})}{4}

Explanation:

Mid point acceleration,
$a = \frac{υ^{2}}{r} = \frac{4}{4} = 1 m s^{- 1}$ at $45^{\circ}$ with $x$ axis
$∴ {\vec{a}}_{x} = - a \cos 45^{\circ} \hat{x}$
$= - 1 \times \frac{1}{\sqrt{2}} \hat{x} = - \frac{1}{\sqrt{2}} \hat{x}$
${\vec{a}}_{y} = - a \sin 45^{\circ} \hat{y}$
$= - 1 \times \frac{1}{\sqrt{2}} \hat{y} = - \frac{1}{\sqrt{2}} \hat{y}$
Thus $\hat{a} = {\vec{a}}_{x} + {\vec{a}}_{y} = - \frac{1}{\sqrt{2}} (\hat{x} + \hat{y})$
supporting img

PHXI04:MOTION IN A PLANE

361886 A ball moves along an uneven horizontal road with fixed speed at all points on the road. The normal reaction of the road on the body at two different points $A$ and $B$ is
supporting img

1 Maximum at

A

2 Maximum at

B

3 Minimum at

B

4 Same at

A

and

B

Explanation:

Centripetal acceleration at $A$ and $B$ is
$\frac{m u^{2}}{r_{A}}$ and $\frac{m u^{2}}{r_{B}}$ respectively.
Clearly $r_{B} > r_{A}$ therefore centripetal acceleration at $A$ is more. Thus, reaction at $B$ is less than $A$ because here the normal reaction provides the centripetal force.

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