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PHXI04:MOTION IN A PLANE

361887 The angle turned by a body undergoing circular motion depends on time as $θ = θ_{0} + θ_{1} t^{2} .$ Then the angular acceleration of the body is

1

θ_{2}

2

2 θ_{0}

3

2 θ_{1}

4

θ_{1}

Explanation:

Angular acceleration $= \frac{d^{2} θ}{d t^{2}} = 2 θ_{1}$

PHXI04:MOTION IN A PLANE

361888 The position vector of a particle is $\vec{r} = (a \cos ω t) \hat{i} + (a \sin ω t) \hat{j}$ . The velocity of the particle is

1 directed towards the origin

2 directed away from the origin

3 parallel to the position vector

4 perpendicular to the position vector

Explanation:

$\vec{r} = (a \cos ω t) \hat{i} + (a \sin ω t) \hat{j}$
$\vec{v} = \frac{d (\vec{r})}{d t} = \frac{d}{d t} {(a \cos ω t) \hat{i} + (a \sin ω t) \hat{j}}$
$= (- a ω \sin ω t) \hat{i} + (a ω \cos ω t) \hat{j}$
$= ω [(- a \sin ω t) \hat{i} + (a \cos ω t) \hat{j}]$
Slope of position vector $= \frac{a \sin ω t}{a \cos ω t} = \tan ω t$
and slope of velocity vector
$= \frac{- a \cos ω t}{a \sin ω t} = \frac{- 1}{\tan ω t}$
$∴$ velocity is perpendicular to the displacement.

PHXI04:MOTION IN A PLANE

361889 A body of mass $m$ is moving a circle of radius $r$ with a constant speed $v$ . The force on the body is $\frac{m v^{2}}{r}$ and is directed towards the centre.
What is the work done by this force in moving the body over half the circumference of the circle?

1

\frac{m v^{2}}{r} \times π r

2 Zero

3

\frac{m v^{2}}{r^{2}}

4

\frac{π r^{2}}{m v^{2}}

PHXI04:MOTION IN A PLANE

361890 A body is travelling in a circle at a constant speed. It

1 has a constant velocity

2 is not accelerated

3 has an inward radial acceleration

4 has an outward radial acceleration

Explanation:

Body moves with constant speed it means that tangential acceleration $a_{T} = 0$ & only centripetal acceleration $a_{C}$ exists whose direction is always towards the centre or inward (along the radius of the circle).

PHXI04:MOTION IN A PLANE

361887 The angle turned by a body undergoing circular motion depends on time as $θ = θ_{0} + θ_{1} t^{2} .$ Then the angular acceleration of the body is

1

θ_{2}

2

2 θ_{0}

3

2 θ_{1}

4

θ_{1}

Explanation:

Angular acceleration $= \frac{d^{2} θ}{d t^{2}} = 2 θ_{1}$

PHXI04:MOTION IN A PLANE

361888 The position vector of a particle is $\vec{r} = (a \cos ω t) \hat{i} + (a \sin ω t) \hat{j}$ . The velocity of the particle is

1 directed towards the origin

2 directed away from the origin

3 parallel to the position vector

4 perpendicular to the position vector

Explanation:

$\vec{r} = (a \cos ω t) \hat{i} + (a \sin ω t) \hat{j}$
$\vec{v} = \frac{d (\vec{r})}{d t} = \frac{d}{d t} {(a \cos ω t) \hat{i} + (a \sin ω t) \hat{j}}$
$= (- a ω \sin ω t) \hat{i} + (a ω \cos ω t) \hat{j}$
$= ω [(- a \sin ω t) \hat{i} + (a \cos ω t) \hat{j}]$
Slope of position vector $= \frac{a \sin ω t}{a \cos ω t} = \tan ω t$
and slope of velocity vector
$= \frac{- a \cos ω t}{a \sin ω t} = \frac{- 1}{\tan ω t}$
$∴$ velocity is perpendicular to the displacement.

PHXI04:MOTION IN A PLANE

361889 A body of mass $m$ is moving a circle of radius $r$ with a constant speed $v$ . The force on the body is $\frac{m v^{2}}{r}$ and is directed towards the centre.
What is the work done by this force in moving the body over half the circumference of the circle?

1

\frac{m v^{2}}{r} \times π r

2 Zero

3

\frac{m v^{2}}{r^{2}}

4

\frac{π r^{2}}{m v^{2}}

PHXI04:MOTION IN A PLANE

361890 A body is travelling in a circle at a constant speed. It

1 has a constant velocity

2 is not accelerated

3 has an inward radial acceleration

4 has an outward radial acceleration

Explanation:

Body moves with constant speed it means that tangential acceleration $a_{T} = 0$ & only centripetal acceleration $a_{C}$ exists whose direction is always towards the centre or inward (along the radius of the circle).

PHXI04:MOTION IN A PLANE

361887 The angle turned by a body undergoing circular motion depends on time as $θ = θ_{0} + θ_{1} t^{2} .$ Then the angular acceleration of the body is

1

θ_{2}

2

2 θ_{0}

3

2 θ_{1}

4

θ_{1}

Explanation:

Angular acceleration $= \frac{d^{2} θ}{d t^{2}} = 2 θ_{1}$

PHXI04:MOTION IN A PLANE

361888 The position vector of a particle is $\vec{r} = (a \cos ω t) \hat{i} + (a \sin ω t) \hat{j}$ . The velocity of the particle is

1 directed towards the origin

2 directed away from the origin

3 parallel to the position vector

4 perpendicular to the position vector

Explanation:

$\vec{r} = (a \cos ω t) \hat{i} + (a \sin ω t) \hat{j}$
$\vec{v} = \frac{d (\vec{r})}{d t} = \frac{d}{d t} {(a \cos ω t) \hat{i} + (a \sin ω t) \hat{j}}$
$= (- a ω \sin ω t) \hat{i} + (a ω \cos ω t) \hat{j}$
$= ω [(- a \sin ω t) \hat{i} + (a \cos ω t) \hat{j}]$
Slope of position vector $= \frac{a \sin ω t}{a \cos ω t} = \tan ω t$
and slope of velocity vector
$= \frac{- a \cos ω t}{a \sin ω t} = \frac{- 1}{\tan ω t}$
$∴$ velocity is perpendicular to the displacement.

PHXI04:MOTION IN A PLANE

361889 A body of mass $m$ is moving a circle of radius $r$ with a constant speed $v$ . The force on the body is $\frac{m v^{2}}{r}$ and is directed towards the centre.
What is the work done by this force in moving the body over half the circumference of the circle?

1

\frac{m v^{2}}{r} \times π r

2 Zero

3

\frac{m v^{2}}{r^{2}}

4

\frac{π r^{2}}{m v^{2}}

PHXI04:MOTION IN A PLANE

361890 A body is travelling in a circle at a constant speed. It

1 has a constant velocity

2 is not accelerated

3 has an inward radial acceleration

4 has an outward radial acceleration

Explanation:

Body moves with constant speed it means that tangential acceleration $a_{T} = 0$ & only centripetal acceleration $a_{C}$ exists whose direction is always towards the centre or inward (along the radius of the circle).

PHXI04:MOTION IN A PLANE

361887 The angle turned by a body undergoing circular motion depends on time as $θ = θ_{0} + θ_{1} t^{2} .$ Then the angular acceleration of the body is

1

θ_{2}

2

2 θ_{0}

3

2 θ_{1}

4

θ_{1}

Explanation:

Angular acceleration $= \frac{d^{2} θ}{d t^{2}} = 2 θ_{1}$

PHXI04:MOTION IN A PLANE

361888 The position vector of a particle is $\vec{r} = (a \cos ω t) \hat{i} + (a \sin ω t) \hat{j}$ . The velocity of the particle is

1 directed towards the origin

2 directed away from the origin

3 parallel to the position vector

4 perpendicular to the position vector

Explanation:

$\vec{r} = (a \cos ω t) \hat{i} + (a \sin ω t) \hat{j}$
$\vec{v} = \frac{d (\vec{r})}{d t} = \frac{d}{d t} {(a \cos ω t) \hat{i} + (a \sin ω t) \hat{j}}$
$= (- a ω \sin ω t) \hat{i} + (a ω \cos ω t) \hat{j}$
$= ω [(- a \sin ω t) \hat{i} + (a \cos ω t) \hat{j}]$
Slope of position vector $= \frac{a \sin ω t}{a \cos ω t} = \tan ω t$
and slope of velocity vector
$= \frac{- a \cos ω t}{a \sin ω t} = \frac{- 1}{\tan ω t}$
$∴$ velocity is perpendicular to the displacement.

PHXI04:MOTION IN A PLANE

361889 A body of mass $m$ is moving a circle of radius $r$ with a constant speed $v$ . The force on the body is $\frac{m v^{2}}{r}$ and is directed towards the centre.
What is the work done by this force in moving the body over half the circumference of the circle?

1

\frac{m v^{2}}{r} \times π r

2 Zero

3

\frac{m v^{2}}{r^{2}}

4

\frac{π r^{2}}{m v^{2}}

PHXI04:MOTION IN A PLANE

361890 A body is travelling in a circle at a constant speed. It

1 has a constant velocity

2 is not accelerated

3 has an inward radial acceleration

4 has an outward radial acceleration

Explanation:

Body moves with constant speed it means that tangential acceleration $a_{T} = 0$ & only centripetal acceleration $a_{C}$ exists whose direction is always towards the centre or inward (along the radius of the circle).

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