361887
The angle turned by a body undergoing circular motion depends on time as \(\theta = {\theta _0} + {\theta _1}{t^2}.\) Then the angular acceleration of the body is
361888
The position vector of a particle is \(\vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}\). 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:
\(\vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}\) \(\vec v = \frac{{d(\vec r)}}{{dt}} = \frac{d}{{dt}}\{ (a\cos \omega t)\hat i + (a\sin \omega t)\hat j\} \) \( = ( - a\omega \sin \omega t)\hat i + (a\omega \cos \omega t)\hat j\) \( = \omega [( - a\sin \omega t)\hat i + (a\cos \omega t)\hat j]\) Slope of position vector \(=\dfrac{a \sin \omega t}{a \cos \omega t}=\tan \omega t\) and slope of velocity vector \(=\dfrac{-a \cos \omega t}{a \sin \omega t}=\dfrac{-1}{\tan \omega t}\) \(\therefore\) 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 \(\dfrac{m v^{2}}{r}\) 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 \(\dfrac{m v^{2}}{r} \times \pi r\)
2 Zero
3 \(\dfrac{m v^{2}}{r^{2}}\)
4 \(\dfrac{\pi r^{2}}{m v^{2}}\)
Explanation:
Work done by centripetal force is always zero.
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).
361887
The angle turned by a body undergoing circular motion depends on time as \(\theta = {\theta _0} + {\theta _1}{t^2}.\) Then the angular acceleration of the body is
361888
The position vector of a particle is \(\vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}\). 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:
\(\vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}\) \(\vec v = \frac{{d(\vec r)}}{{dt}} = \frac{d}{{dt}}\{ (a\cos \omega t)\hat i + (a\sin \omega t)\hat j\} \) \( = ( - a\omega \sin \omega t)\hat i + (a\omega \cos \omega t)\hat j\) \( = \omega [( - a\sin \omega t)\hat i + (a\cos \omega t)\hat j]\) Slope of position vector \(=\dfrac{a \sin \omega t}{a \cos \omega t}=\tan \omega t\) and slope of velocity vector \(=\dfrac{-a \cos \omega t}{a \sin \omega t}=\dfrac{-1}{\tan \omega t}\) \(\therefore\) 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 \(\dfrac{m v^{2}}{r}\) 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 \(\dfrac{m v^{2}}{r} \times \pi r\)
2 Zero
3 \(\dfrac{m v^{2}}{r^{2}}\)
4 \(\dfrac{\pi r^{2}}{m v^{2}}\)
Explanation:
Work done by centripetal force is always zero.
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).
361887
The angle turned by a body undergoing circular motion depends on time as \(\theta = {\theta _0} + {\theta _1}{t^2}.\) Then the angular acceleration of the body is
361888
The position vector of a particle is \(\vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}\). 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:
\(\vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}\) \(\vec v = \frac{{d(\vec r)}}{{dt}} = \frac{d}{{dt}}\{ (a\cos \omega t)\hat i + (a\sin \omega t)\hat j\} \) \( = ( - a\omega \sin \omega t)\hat i + (a\omega \cos \omega t)\hat j\) \( = \omega [( - a\sin \omega t)\hat i + (a\cos \omega t)\hat j]\) Slope of position vector \(=\dfrac{a \sin \omega t}{a \cos \omega t}=\tan \omega t\) and slope of velocity vector \(=\dfrac{-a \cos \omega t}{a \sin \omega t}=\dfrac{-1}{\tan \omega t}\) \(\therefore\) 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 \(\dfrac{m v^{2}}{r}\) 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 \(\dfrac{m v^{2}}{r} \times \pi r\)
2 Zero
3 \(\dfrac{m v^{2}}{r^{2}}\)
4 \(\dfrac{\pi r^{2}}{m v^{2}}\)
Explanation:
Work done by centripetal force is always zero.
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).
361887
The angle turned by a body undergoing circular motion depends on time as \(\theta = {\theta _0} + {\theta _1}{t^2}.\) Then the angular acceleration of the body is
361888
The position vector of a particle is \(\vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}\). 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:
\(\vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}\) \(\vec v = \frac{{d(\vec r)}}{{dt}} = \frac{d}{{dt}}\{ (a\cos \omega t)\hat i + (a\sin \omega t)\hat j\} \) \( = ( - a\omega \sin \omega t)\hat i + (a\omega \cos \omega t)\hat j\) \( = \omega [( - a\sin \omega t)\hat i + (a\cos \omega t)\hat j]\) Slope of position vector \(=\dfrac{a \sin \omega t}{a \cos \omega t}=\tan \omega t\) and slope of velocity vector \(=\dfrac{-a \cos \omega t}{a \sin \omega t}=\dfrac{-1}{\tan \omega t}\) \(\therefore\) 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 \(\dfrac{m v^{2}}{r}\) 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 \(\dfrac{m v^{2}}{r} \times \pi r\)
2 Zero
3 \(\dfrac{m v^{2}}{r^{2}}\)
4 \(\dfrac{\pi r^{2}}{m v^{2}}\)
Explanation:
Work done by centripetal force is always zero.
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).
