361861
A particle completes 3 revolutions per second on a circular path of radius \(8\;cm\). Find the value of angular velocity and centripetal acceleration of the particle
To find the angular velocity \((\omega)\) and centripetal acceleration \(\left(a_{r}\right)\) of a particle moving in a circle with 3 revolutions per second (frequency \(f = 3\;Hz\)) and radius \(r = 8\;cm\). Angular velocity \(\omega\) is given by \(\omega=2 \pi f\), so \(\omega = 2\pi \times 3\,rad{\rm{/}}s = 6\pi \,rad{\rm{/}}s\). Centripetal acceleration \(a_{r}\) is given by \(a_{r}=\omega^{2} r\), so \({a_r} = {(6\pi )^2} \times 8 = 288{\pi ^2}\;cm{\rm{/}}{s^2}\) Therefore, the correct option is (1).
PHXI04:MOTION IN A PLANE
361862
If \(a_{r}\) and \(a_{t}\) represent radial and tangential accelerations respectively, the motion of a particle will be uniformly circular if
1 \(a_{r}=0\) and \(a_{t}=0\)
2 \(a_{r}=0\) but \(a_{t} \neq 0\)
3 \(a_{r} \neq 0\) but \(a_{t}=0\)
4 \(a_{r} \neq 0\) and \(a_{t} \neq 0\)
Explanation:
If \(a_{r} \neq 0\) but \(a_{t}=0\), then motion is a uniform circular. If \(a_{r} \neq 0\) and \(a_{t} \neq 0\), then motion is a non- uniform circular.
PHXI04:MOTION IN A PLANE
361863
Assertion : A uniform circular motion is an accelerated motion. Reason : Direction of acceleration is parallel to the velocity vector.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The acceleration experienced by an object undergoing uniform circular motion is known as centripetal acceleration, and it always points toward the center of the circular path. So correct option is (3).
PHXI04:MOTION IN A PLANE
361864
A Particle moves along a circle of radius \(r\) with constant tangential acceleration. If the velocity of the particle is \(v\) at the end of second revolution, after the revolution has started, then the tangential acceleration is
1 \(\frac{{{v^2}}}{{8\pi r}}\)
2 \(\frac{{{v^2}}}{{6\pi r}}\)
3 \(\frac{{{v^2}}}{{4\pi r}}\)
4 \(\frac{{{v^2}}}{{2\pi r}}\)
Explanation:
Applying equation of motion in tangential direction (along the circular path), \({v^2} = {u^2} + 2as\) [Symbols have their usual meanings] Given, \(u = 0\) and \(s = 2 \times 2\pi r = 4\pi r\) \(\therefore \,\,{v^2} = 2a \times 4\pi r \Rightarrow a = \frac{{{v^2}}}{{8\pi r}}\)
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PHXI04:MOTION IN A PLANE
361861
A particle completes 3 revolutions per second on a circular path of radius \(8\;cm\). Find the value of angular velocity and centripetal acceleration of the particle
To find the angular velocity \((\omega)\) and centripetal acceleration \(\left(a_{r}\right)\) of a particle moving in a circle with 3 revolutions per second (frequency \(f = 3\;Hz\)) and radius \(r = 8\;cm\). Angular velocity \(\omega\) is given by \(\omega=2 \pi f\), so \(\omega = 2\pi \times 3\,rad{\rm{/}}s = 6\pi \,rad{\rm{/}}s\). Centripetal acceleration \(a_{r}\) is given by \(a_{r}=\omega^{2} r\), so \({a_r} = {(6\pi )^2} \times 8 = 288{\pi ^2}\;cm{\rm{/}}{s^2}\) Therefore, the correct option is (1).
PHXI04:MOTION IN A PLANE
361862
If \(a_{r}\) and \(a_{t}\) represent radial and tangential accelerations respectively, the motion of a particle will be uniformly circular if
1 \(a_{r}=0\) and \(a_{t}=0\)
2 \(a_{r}=0\) but \(a_{t} \neq 0\)
3 \(a_{r} \neq 0\) but \(a_{t}=0\)
4 \(a_{r} \neq 0\) and \(a_{t} \neq 0\)
Explanation:
If \(a_{r} \neq 0\) but \(a_{t}=0\), then motion is a uniform circular. If \(a_{r} \neq 0\) and \(a_{t} \neq 0\), then motion is a non- uniform circular.
PHXI04:MOTION IN A PLANE
361863
Assertion : A uniform circular motion is an accelerated motion. Reason : Direction of acceleration is parallel to the velocity vector.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The acceleration experienced by an object undergoing uniform circular motion is known as centripetal acceleration, and it always points toward the center of the circular path. So correct option is (3).
PHXI04:MOTION IN A PLANE
361864
A Particle moves along a circle of radius \(r\) with constant tangential acceleration. If the velocity of the particle is \(v\) at the end of second revolution, after the revolution has started, then the tangential acceleration is
1 \(\frac{{{v^2}}}{{8\pi r}}\)
2 \(\frac{{{v^2}}}{{6\pi r}}\)
3 \(\frac{{{v^2}}}{{4\pi r}}\)
4 \(\frac{{{v^2}}}{{2\pi r}}\)
Explanation:
Applying equation of motion in tangential direction (along the circular path), \({v^2} = {u^2} + 2as\) [Symbols have their usual meanings] Given, \(u = 0\) and \(s = 2 \times 2\pi r = 4\pi r\) \(\therefore \,\,{v^2} = 2a \times 4\pi r \Rightarrow a = \frac{{{v^2}}}{{8\pi r}}\)
361861
A particle completes 3 revolutions per second on a circular path of radius \(8\;cm\). Find the value of angular velocity and centripetal acceleration of the particle
To find the angular velocity \((\omega)\) and centripetal acceleration \(\left(a_{r}\right)\) of a particle moving in a circle with 3 revolutions per second (frequency \(f = 3\;Hz\)) and radius \(r = 8\;cm\). Angular velocity \(\omega\) is given by \(\omega=2 \pi f\), so \(\omega = 2\pi \times 3\,rad{\rm{/}}s = 6\pi \,rad{\rm{/}}s\). Centripetal acceleration \(a_{r}\) is given by \(a_{r}=\omega^{2} r\), so \({a_r} = {(6\pi )^2} \times 8 = 288{\pi ^2}\;cm{\rm{/}}{s^2}\) Therefore, the correct option is (1).
PHXI04:MOTION IN A PLANE
361862
If \(a_{r}\) and \(a_{t}\) represent radial and tangential accelerations respectively, the motion of a particle will be uniformly circular if
1 \(a_{r}=0\) and \(a_{t}=0\)
2 \(a_{r}=0\) but \(a_{t} \neq 0\)
3 \(a_{r} \neq 0\) but \(a_{t}=0\)
4 \(a_{r} \neq 0\) and \(a_{t} \neq 0\)
Explanation:
If \(a_{r} \neq 0\) but \(a_{t}=0\), then motion is a uniform circular. If \(a_{r} \neq 0\) and \(a_{t} \neq 0\), then motion is a non- uniform circular.
PHXI04:MOTION IN A PLANE
361863
Assertion : A uniform circular motion is an accelerated motion. Reason : Direction of acceleration is parallel to the velocity vector.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The acceleration experienced by an object undergoing uniform circular motion is known as centripetal acceleration, and it always points toward the center of the circular path. So correct option is (3).
PHXI04:MOTION IN A PLANE
361864
A Particle moves along a circle of radius \(r\) with constant tangential acceleration. If the velocity of the particle is \(v\) at the end of second revolution, after the revolution has started, then the tangential acceleration is
1 \(\frac{{{v^2}}}{{8\pi r}}\)
2 \(\frac{{{v^2}}}{{6\pi r}}\)
3 \(\frac{{{v^2}}}{{4\pi r}}\)
4 \(\frac{{{v^2}}}{{2\pi r}}\)
Explanation:
Applying equation of motion in tangential direction (along the circular path), \({v^2} = {u^2} + 2as\) [Symbols have their usual meanings] Given, \(u = 0\) and \(s = 2 \times 2\pi r = 4\pi r\) \(\therefore \,\,{v^2} = 2a \times 4\pi r \Rightarrow a = \frac{{{v^2}}}{{8\pi r}}\)
361861
A particle completes 3 revolutions per second on a circular path of radius \(8\;cm\). Find the value of angular velocity and centripetal acceleration of the particle
To find the angular velocity \((\omega)\) and centripetal acceleration \(\left(a_{r}\right)\) of a particle moving in a circle with 3 revolutions per second (frequency \(f = 3\;Hz\)) and radius \(r = 8\;cm\). Angular velocity \(\omega\) is given by \(\omega=2 \pi f\), so \(\omega = 2\pi \times 3\,rad{\rm{/}}s = 6\pi \,rad{\rm{/}}s\). Centripetal acceleration \(a_{r}\) is given by \(a_{r}=\omega^{2} r\), so \({a_r} = {(6\pi )^2} \times 8 = 288{\pi ^2}\;cm{\rm{/}}{s^2}\) Therefore, the correct option is (1).
PHXI04:MOTION IN A PLANE
361862
If \(a_{r}\) and \(a_{t}\) represent radial and tangential accelerations respectively, the motion of a particle will be uniformly circular if
1 \(a_{r}=0\) and \(a_{t}=0\)
2 \(a_{r}=0\) but \(a_{t} \neq 0\)
3 \(a_{r} \neq 0\) but \(a_{t}=0\)
4 \(a_{r} \neq 0\) and \(a_{t} \neq 0\)
Explanation:
If \(a_{r} \neq 0\) but \(a_{t}=0\), then motion is a uniform circular. If \(a_{r} \neq 0\) and \(a_{t} \neq 0\), then motion is a non- uniform circular.
PHXI04:MOTION IN A PLANE
361863
Assertion : A uniform circular motion is an accelerated motion. Reason : Direction of acceleration is parallel to the velocity vector.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The acceleration experienced by an object undergoing uniform circular motion is known as centripetal acceleration, and it always points toward the center of the circular path. So correct option is (3).
PHXI04:MOTION IN A PLANE
361864
A Particle moves along a circle of radius \(r\) with constant tangential acceleration. If the velocity of the particle is \(v\) at the end of second revolution, after the revolution has started, then the tangential acceleration is
1 \(\frac{{{v^2}}}{{8\pi r}}\)
2 \(\frac{{{v^2}}}{{6\pi r}}\)
3 \(\frac{{{v^2}}}{{4\pi r}}\)
4 \(\frac{{{v^2}}}{{2\pi r}}\)
Explanation:
Applying equation of motion in tangential direction (along the circular path), \({v^2} = {u^2} + 2as\) [Symbols have their usual meanings] Given, \(u = 0\) and \(s = 2 \times 2\pi r = 4\pi r\) \(\therefore \,\,{v^2} = 2a \times 4\pi r \Rightarrow a = \frac{{{v^2}}}{{8\pi r}}\)