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A particle of mass 500 gm is at rest in the stationary reference frame. The motion of the particle with respect to reference frame rotating with an angular velocity 10π rad/sec2 is.
- a)elliptical
- b)eqvi angular
- c)circular
- d)parabolic
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A particle of mass 500 gm is at rest in the stationary reference frame...
Mass of particle m = 500 gm = 0.5 kg angular velocity of reference frame ω = 10π rad/sec2 Distance of particle from the axis of rotation r= 15 cm = 0.15 m
We know that the velocity of particle in rotating reference frame
Since the particle is at rest in the stationary reference frame
so, the particle will appear to be moving in a circular path with respect to rotating reference frame.
We know that the velocity of particle in rotating reference frame
Since the particle is at rest in the stationary reference frame
so, the particle will appear to be moving in a circular path with respect to rotating reference frame.
Most Upvoted Answer
A particle of mass 500 gm is at rest in the stationary reference frame...
Rad/s about the z-axis is described by the equation of motion:
r(t) = (2cos(10t), 2sin(10t), 0)
where r is the position vector of the particle and t is time.
To find the velocity and acceleration of the particle, we need to take the first and second derivatives of the position vector with respect to time.
Velocity:
v(t) = dr/dt = (-4sin(10t), 4cos(10t), 0)
Acceleration:
a(t) = d^2r/dt^2 = (-40cos(10t), -40sin(10t), 0)
In the stationary reference frame, the particle is at rest, so its velocity and acceleration are both zero. However, in the rotating reference frame, the particle has a non-zero velocity and acceleration due to the rotation of the reference frame.
The velocity of the particle in the rotating reference frame can be found by subtracting the velocity of the rotating reference frame from the velocity of the particle in the stationary reference frame:
v' = v - r x omega
where omega is the angular velocity of the rotating reference frame and x represents the cross product.
Substituting the given values, we get:
v' = (-4sin(10t), 4cos(10t), 0) - (0, 0, 20)
Simplifying, we get:
v' = (-4sin(10t), 4cos(10t), -20)
Similarly, the acceleration of the particle in the rotating reference frame can be found by subtracting the acceleration of the rotating reference frame from the acceleration of the particle in the stationary reference frame:
a' = a - 2v x omega - r x omega^2
Substituting the given values, we get:
a' = (-40cos(10t), -40sin(10t), 0) - (0, 0, 40sin(10t)) - (0, 0, -200)
Simplifying, we get:
a' = (-40cos(10t), -40sin(10t), 200 - 40sin(10t))
Therefore, the velocity and acceleration of the particle in the rotating reference frame are given by:
v' = (-4sin(10t), 4cos(10t), -20)
a' = (-40cos(10t), -40sin(10t), 200 - 40sin(10t))
r(t) = (2cos(10t), 2sin(10t), 0)
where r is the position vector of the particle and t is time.
To find the velocity and acceleration of the particle, we need to take the first and second derivatives of the position vector with respect to time.
Velocity:
v(t) = dr/dt = (-4sin(10t), 4cos(10t), 0)
Acceleration:
a(t) = d^2r/dt^2 = (-40cos(10t), -40sin(10t), 0)
In the stationary reference frame, the particle is at rest, so its velocity and acceleration are both zero. However, in the rotating reference frame, the particle has a non-zero velocity and acceleration due to the rotation of the reference frame.
The velocity of the particle in the rotating reference frame can be found by subtracting the velocity of the rotating reference frame from the velocity of the particle in the stationary reference frame:
v' = v - r x omega
where omega is the angular velocity of the rotating reference frame and x represents the cross product.
Substituting the given values, we get:
v' = (-4sin(10t), 4cos(10t), 0) - (0, 0, 20)
Simplifying, we get:
v' = (-4sin(10t), 4cos(10t), -20)
Similarly, the acceleration of the particle in the rotating reference frame can be found by subtracting the acceleration of the rotating reference frame from the acceleration of the particle in the stationary reference frame:
a' = a - 2v x omega - r x omega^2
Substituting the given values, we get:
a' = (-40cos(10t), -40sin(10t), 0) - (0, 0, 40sin(10t)) - (0, 0, -200)
Simplifying, we get:
a' = (-40cos(10t), -40sin(10t), 200 - 40sin(10t))
Therefore, the velocity and acceleration of the particle in the rotating reference frame are given by:
v' = (-4sin(10t), 4cos(10t), -20)
a' = (-40cos(10t), -40sin(10t), 200 - 40sin(10t))
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A particle of mass 500 gm is at rest in the stationary reference frame. The motion of the particle with respect to reference frame rotating with an angular velocity 10π rad/sec2 is.a)ellipticalb)eqvi angularc)circulard)parabolicCorrect answer is option 'C'. Can you explain this answer?
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