361848
An object moves at a constant speed along a circular path in a horizontal \(XY\) plane, with the centre at the origin. When the object is at \(x = - 2m\), its velocity is \( - (4m/s)\widehat j\). What is the object’s acceleration when it is \(y = 2m\)?
1 \( - (8m/{s^2})\widehat j\)
2 \( - (8m/{s^2})\widehat i\)
3 \( - (4m/{s^2})\widehat j\)
4 \( - (4m/{s^2})\widehat i\)
Explanation:
The radius of circular path is 2\(m\) and the speed of the object is 4\(m\)/\(s\). Acceleration of the object is shown at different positions The magnitude of acceleration is \(a = \frac{{{v^2}}}{R} = \frac{{{{\left( 4 \right)}^2}}}{2} = 8\,m/{s^2}\) The acceleration is directed towards the centre. Therefore, when an object is at \(y = 2m\), its acceleration is \( - 8\widehat j\,m/{s^2}\)
PHXI04:MOTION IN A PLANE
361849
The ratio of angular speeds of minute hand and hour hand of a watch is
361850
Which one of the following statements is not correct in uniform circular motion?
1 The speed of the particle remains constant.
2 The acceleration always points towards the centre.
3 The angular speed remains constant.
4 The velocity remains constant.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361851
The magnitude of the displacement of a particle moving in a circle of radius \(a\) with constant angular speed \(\omega \) varies with time \(t\) as:
1 \(2a\sin \omega t\)
2 \(2a\sin \frac{{\omega t}}{2}\)
3 \(2a\cos \omega t\)
4 \(2a\cos \frac{{\omega t}}{2}\)
Explanation:
In time \(t\), particle has rotated an angle \(\theta = \omega t\) The net displacement is \(s = \sqrt {{y^2} + {{\left( {a - x} \right)}^2}} \) \(y = a\sin \theta = a\sin \omega t\) \(x = a\cos \theta = a\cos \omega t\) \( \Rightarrow s = \sqrt {{{\left( {a\sin \omega t} \right)}^2} + {{\left( {a - a\cos \omega t} \right)}^2}} \) \( \Rightarrow s = 2a\sin \frac{{\omega t}}{2}\)
PHXI04:MOTION IN A PLANE
361852
A scooter is going round a circular road of radius 200 \(m\) at a speed of \({20 {~m} / {s}}\). The angular speed of scooter will be
1 1 radian/\(\sec \)
2 0.01 radian/\(\sec \)
3 0.1 radian/\(\sec \)
4 10 radian/ \(\sec \)
Explanation:
Radius of the circular road, \(r=200 {~m}\) Speed of scooter, \({v=20 {~m} / {s}}\) Angular speed of scooter will be given by \(v=r \omega \Rightarrow \omega=\dfrac{v}{r}\) \({\Rightarrow \omega=\dfrac{20}{200} {rad} / {s}}\) \( \Rightarrow \omega = 0.1\,\,rad/s\). So correct option is (3)
361848
An object moves at a constant speed along a circular path in a horizontal \(XY\) plane, with the centre at the origin. When the object is at \(x = - 2m\), its velocity is \( - (4m/s)\widehat j\). What is the object’s acceleration when it is \(y = 2m\)?
1 \( - (8m/{s^2})\widehat j\)
2 \( - (8m/{s^2})\widehat i\)
3 \( - (4m/{s^2})\widehat j\)
4 \( - (4m/{s^2})\widehat i\)
Explanation:
The radius of circular path is 2\(m\) and the speed of the object is 4\(m\)/\(s\). Acceleration of the object is shown at different positions The magnitude of acceleration is \(a = \frac{{{v^2}}}{R} = \frac{{{{\left( 4 \right)}^2}}}{2} = 8\,m/{s^2}\) The acceleration is directed towards the centre. Therefore, when an object is at \(y = 2m\), its acceleration is \( - 8\widehat j\,m/{s^2}\)
PHXI04:MOTION IN A PLANE
361849
The ratio of angular speeds of minute hand and hour hand of a watch is
361850
Which one of the following statements is not correct in uniform circular motion?
1 The speed of the particle remains constant.
2 The acceleration always points towards the centre.
3 The angular speed remains constant.
4 The velocity remains constant.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361851
The magnitude of the displacement of a particle moving in a circle of radius \(a\) with constant angular speed \(\omega \) varies with time \(t\) as:
1 \(2a\sin \omega t\)
2 \(2a\sin \frac{{\omega t}}{2}\)
3 \(2a\cos \omega t\)
4 \(2a\cos \frac{{\omega t}}{2}\)
Explanation:
In time \(t\), particle has rotated an angle \(\theta = \omega t\) The net displacement is \(s = \sqrt {{y^2} + {{\left( {a - x} \right)}^2}} \) \(y = a\sin \theta = a\sin \omega t\) \(x = a\cos \theta = a\cos \omega t\) \( \Rightarrow s = \sqrt {{{\left( {a\sin \omega t} \right)}^2} + {{\left( {a - a\cos \omega t} \right)}^2}} \) \( \Rightarrow s = 2a\sin \frac{{\omega t}}{2}\)
PHXI04:MOTION IN A PLANE
361852
A scooter is going round a circular road of radius 200 \(m\) at a speed of \({20 {~m} / {s}}\). The angular speed of scooter will be
1 1 radian/\(\sec \)
2 0.01 radian/\(\sec \)
3 0.1 radian/\(\sec \)
4 10 radian/ \(\sec \)
Explanation:
Radius of the circular road, \(r=200 {~m}\) Speed of scooter, \({v=20 {~m} / {s}}\) Angular speed of scooter will be given by \(v=r \omega \Rightarrow \omega=\dfrac{v}{r}\) \({\Rightarrow \omega=\dfrac{20}{200} {rad} / {s}}\) \( \Rightarrow \omega = 0.1\,\,rad/s\). So correct option is (3)
361848
An object moves at a constant speed along a circular path in a horizontal \(XY\) plane, with the centre at the origin. When the object is at \(x = - 2m\), its velocity is \( - (4m/s)\widehat j\). What is the object’s acceleration when it is \(y = 2m\)?
1 \( - (8m/{s^2})\widehat j\)
2 \( - (8m/{s^2})\widehat i\)
3 \( - (4m/{s^2})\widehat j\)
4 \( - (4m/{s^2})\widehat i\)
Explanation:
The radius of circular path is 2\(m\) and the speed of the object is 4\(m\)/\(s\). Acceleration of the object is shown at different positions The magnitude of acceleration is \(a = \frac{{{v^2}}}{R} = \frac{{{{\left( 4 \right)}^2}}}{2} = 8\,m/{s^2}\) The acceleration is directed towards the centre. Therefore, when an object is at \(y = 2m\), its acceleration is \( - 8\widehat j\,m/{s^2}\)
PHXI04:MOTION IN A PLANE
361849
The ratio of angular speeds of minute hand and hour hand of a watch is
361850
Which one of the following statements is not correct in uniform circular motion?
1 The speed of the particle remains constant.
2 The acceleration always points towards the centre.
3 The angular speed remains constant.
4 The velocity remains constant.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361851
The magnitude of the displacement of a particle moving in a circle of radius \(a\) with constant angular speed \(\omega \) varies with time \(t\) as:
1 \(2a\sin \omega t\)
2 \(2a\sin \frac{{\omega t}}{2}\)
3 \(2a\cos \omega t\)
4 \(2a\cos \frac{{\omega t}}{2}\)
Explanation:
In time \(t\), particle has rotated an angle \(\theta = \omega t\) The net displacement is \(s = \sqrt {{y^2} + {{\left( {a - x} \right)}^2}} \) \(y = a\sin \theta = a\sin \omega t\) \(x = a\cos \theta = a\cos \omega t\) \( \Rightarrow s = \sqrt {{{\left( {a\sin \omega t} \right)}^2} + {{\left( {a - a\cos \omega t} \right)}^2}} \) \( \Rightarrow s = 2a\sin \frac{{\omega t}}{2}\)
PHXI04:MOTION IN A PLANE
361852
A scooter is going round a circular road of radius 200 \(m\) at a speed of \({20 {~m} / {s}}\). The angular speed of scooter will be
1 1 radian/\(\sec \)
2 0.01 radian/\(\sec \)
3 0.1 radian/\(\sec \)
4 10 radian/ \(\sec \)
Explanation:
Radius of the circular road, \(r=200 {~m}\) Speed of scooter, \({v=20 {~m} / {s}}\) Angular speed of scooter will be given by \(v=r \omega \Rightarrow \omega=\dfrac{v}{r}\) \({\Rightarrow \omega=\dfrac{20}{200} {rad} / {s}}\) \( \Rightarrow \omega = 0.1\,\,rad/s\). So correct option is (3)
361848
An object moves at a constant speed along a circular path in a horizontal \(XY\) plane, with the centre at the origin. When the object is at \(x = - 2m\), its velocity is \( - (4m/s)\widehat j\). What is the object’s acceleration when it is \(y = 2m\)?
1 \( - (8m/{s^2})\widehat j\)
2 \( - (8m/{s^2})\widehat i\)
3 \( - (4m/{s^2})\widehat j\)
4 \( - (4m/{s^2})\widehat i\)
Explanation:
The radius of circular path is 2\(m\) and the speed of the object is 4\(m\)/\(s\). Acceleration of the object is shown at different positions The magnitude of acceleration is \(a = \frac{{{v^2}}}{R} = \frac{{{{\left( 4 \right)}^2}}}{2} = 8\,m/{s^2}\) The acceleration is directed towards the centre. Therefore, when an object is at \(y = 2m\), its acceleration is \( - 8\widehat j\,m/{s^2}\)
PHXI04:MOTION IN A PLANE
361849
The ratio of angular speeds of minute hand and hour hand of a watch is
361850
Which one of the following statements is not correct in uniform circular motion?
1 The speed of the particle remains constant.
2 The acceleration always points towards the centre.
3 The angular speed remains constant.
4 The velocity remains constant.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361851
The magnitude of the displacement of a particle moving in a circle of radius \(a\) with constant angular speed \(\omega \) varies with time \(t\) as:
1 \(2a\sin \omega t\)
2 \(2a\sin \frac{{\omega t}}{2}\)
3 \(2a\cos \omega t\)
4 \(2a\cos \frac{{\omega t}}{2}\)
Explanation:
In time \(t\), particle has rotated an angle \(\theta = \omega t\) The net displacement is \(s = \sqrt {{y^2} + {{\left( {a - x} \right)}^2}} \) \(y = a\sin \theta = a\sin \omega t\) \(x = a\cos \theta = a\cos \omega t\) \( \Rightarrow s = \sqrt {{{\left( {a\sin \omega t} \right)}^2} + {{\left( {a - a\cos \omega t} \right)}^2}} \) \( \Rightarrow s = 2a\sin \frac{{\omega t}}{2}\)
PHXI04:MOTION IN A PLANE
361852
A scooter is going round a circular road of radius 200 \(m\) at a speed of \({20 {~m} / {s}}\). The angular speed of scooter will be
1 1 radian/\(\sec \)
2 0.01 radian/\(\sec \)
3 0.1 radian/\(\sec \)
4 10 radian/ \(\sec \)
Explanation:
Radius of the circular road, \(r=200 {~m}\) Speed of scooter, \({v=20 {~m} / {s}}\) Angular speed of scooter will be given by \(v=r \omega \Rightarrow \omega=\dfrac{v}{r}\) \({\Rightarrow \omega=\dfrac{20}{200} {rad} / {s}}\) \( \Rightarrow \omega = 0.1\,\,rad/s\). So correct option is (3)
361848
An object moves at a constant speed along a circular path in a horizontal \(XY\) plane, with the centre at the origin. When the object is at \(x = - 2m\), its velocity is \( - (4m/s)\widehat j\). What is the object’s acceleration when it is \(y = 2m\)?
1 \( - (8m/{s^2})\widehat j\)
2 \( - (8m/{s^2})\widehat i\)
3 \( - (4m/{s^2})\widehat j\)
4 \( - (4m/{s^2})\widehat i\)
Explanation:
The radius of circular path is 2\(m\) and the speed of the object is 4\(m\)/\(s\). Acceleration of the object is shown at different positions The magnitude of acceleration is \(a = \frac{{{v^2}}}{R} = \frac{{{{\left( 4 \right)}^2}}}{2} = 8\,m/{s^2}\) The acceleration is directed towards the centre. Therefore, when an object is at \(y = 2m\), its acceleration is \( - 8\widehat j\,m/{s^2}\)
PHXI04:MOTION IN A PLANE
361849
The ratio of angular speeds of minute hand and hour hand of a watch is
361850
Which one of the following statements is not correct in uniform circular motion?
1 The speed of the particle remains constant.
2 The acceleration always points towards the centre.
3 The angular speed remains constant.
4 The velocity remains constant.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361851
The magnitude of the displacement of a particle moving in a circle of radius \(a\) with constant angular speed \(\omega \) varies with time \(t\) as:
1 \(2a\sin \omega t\)
2 \(2a\sin \frac{{\omega t}}{2}\)
3 \(2a\cos \omega t\)
4 \(2a\cos \frac{{\omega t}}{2}\)
Explanation:
In time \(t\), particle has rotated an angle \(\theta = \omega t\) The net displacement is \(s = \sqrt {{y^2} + {{\left( {a - x} \right)}^2}} \) \(y = a\sin \theta = a\sin \omega t\) \(x = a\cos \theta = a\cos \omega t\) \( \Rightarrow s = \sqrt {{{\left( {a\sin \omega t} \right)}^2} + {{\left( {a - a\cos \omega t} \right)}^2}} \) \( \Rightarrow s = 2a\sin \frac{{\omega t}}{2}\)
PHXI04:MOTION IN A PLANE
361852
A scooter is going round a circular road of radius 200 \(m\) at a speed of \({20 {~m} / {s}}\). The angular speed of scooter will be
1 1 radian/\(\sec \)
2 0.01 radian/\(\sec \)
3 0.1 radian/\(\sec \)
4 10 radian/ \(\sec \)
Explanation:
Radius of the circular road, \(r=200 {~m}\) Speed of scooter, \({v=20 {~m} / {s}}\) Angular speed of scooter will be given by \(v=r \omega \Rightarrow \omega=\dfrac{v}{r}\) \({\Rightarrow \omega=\dfrac{20}{200} {rad} / {s}}\) \( \Rightarrow \omega = 0.1\,\,rad/s\). So correct option is (3)
