Questions OBJECTIVE - I and Answer (Circular Motion) HC Verma Part 1 (1-8)
Q#1
When a particle moves in a circle with a uniform speed(a) Its velocity and acceleration are both constant
(b) Its velocity is constant but the acceleration changes
(c) Its acceleration is constant but the velocity changes
(d) Its velocity and acceleration both change.
Answer: (d)
Velocity and acceleration are both vectors and a vectors is fully defined with its magnitude and direction. Even when one entity changes, the vector changes. In the given case the particle moves in a circle with uniform speed. But movement in a circle makes its direction change every moment. So magnitude of the velocity is constant but its direction changes and hence the velocity changes.
The magnitude of acceleration of the particle is given by v²/r and its direction is towards the center ie towards radius, but as the particle moves the connected radius and hence direction of the acceleration changes. So acceleration also changes as the particle moves.
Q#2
Two cars having masses m1 and m2 move in circles of radii r1 and r2 respectively. If they complete the circles in equal time, the ratio of their angular speeds ω1/ω2 is
(a) m1/m2
(b) r1/r2
(c) m1r1/m2r2
(d) 1
Answer: (d)
In completing a circle the angle covered is 2π what ever be the radius of the circle. The angular speed will be given by 2π/T where T is the time taken in covering the circle. In this case T is same for both cars, so both cars will have same angular speed ω = 2π/T . Hence ω1/ω2 =1
Q#3
A car moves at a constant speed on a road as shown in the figure (7-Q2).The normal force by the road on the car is NA and NB when it is at the points A and B respectively.
(a) NA = NB (b) NA > NB (c) NA < NB (d) Insufficient information to decide the relation of NA and NB.
Answer: (c)
Both points A and B are on the crest of the curves where the weight of the car 'mg' is perpendicular to the surface. But due to the movement of the car on the curve it will feel reduction in the weight by an amount mv²/r, where v is the constant speed of the car and r is the radius of the curve. Clearly this reduction in weight is inversely proportional to r as the numerator is constant. Smaller the r, greater is the term mv²/r. Since r at point A is smaller than at the point B, so reduction in weight will be more at A than at B. It means the apparent weight is less at A than B.
Normal forces at these points will be equal to the apparent weight at these points. So, NA < NB.
Q#4
A particle of mass m is observed from an inertial frame of reference and is found to move in a circle of radius r with a uniform speed v. The centrifugal force on it is,
(a) mv²/r towards the center
(b) mv²/r away from the center
(c) mv²/r along the tangent through the particle.
(d) Zero
Answer: (d)
Centrifugal force is a pseudo force needed in an non-inertial frame of reference when we still want to use the Newton's Laws. In this case the frame of reference is inertial, so there is no centrifugal force,
Q#5
A particle of mass 'm' rotates in a circle of radius 'a' with a uniform angular speed ω. It is viewed from a frame rotating about Z-axis with a uniform angular speed ω0 . The centrifugal force on the particle is
(a) mω²a (b) mω0²a (c) m{(ω + ω0)/2}²a (d) mωω0a
Answer: (b)
The frame of reference is rotating with a uniform angular speed ω0 , so it has an acceleration towards the center and it is a non-inertial plane. To apply Newton's Laws in this frame we apply a pseudo force equal to mω0²a to the particle but opposite to the direction of the acceleration of the frame. This pseudo force is the centrifugal force on the particle.
Q#6
A particle is kept fixed on a turn table rotating uniformly. As seen from the ground, the particle goes in a circle, its speed is 20 cm/s and acceleration 20 cm/s². The particle is now shifted to a new position to make the radius half of the original value. The new values of the speed and the acceleration will be(a) 10 cm/s, 10 cm/s² (b) 10 cm/s, 80 cm/s²
(a) 40 cm/s, 10 cm/s² (a) 40 cm/s, 40 cm/s²
Answer: (a)
Velocity of the particle in a circular motion is equal to ωr, so if ω is constant the velocity is proportional to the radius. Here the particle is kept at half radius therefore velocity will be half ie = 10 cm/s.
Similarly acceleration is ω²r , here too, ω² is constant, so acceleration is also proportional to r. In this case too, making radius half will make acceleration half ie = 10 cm/s².
Q#7
Water in a bucket is whirled in a vertical circle with a string attached to to it. The water does not fall down even when the bucket is inverted at the top of its path. We conclude that in this position
(a) mg = mv²/r (b) mg is greater than mv²/r
(c) mg is not greater than mv²/r
(d) mg is not less than mv²/r
Answer: (c)
At the top position the weight of the water mg acts downward to bring it down while due to inertia water tries to escape upward with a force mv²/r on the bucket. Therefore if the water is not falling at the top position it means mg < mv²/r.
Q#8
A stone of mass m tied to a string of length l is rotated in a circle with the other end of the string as the center. The speed of the stone is v. If the string breaks, the stone will move
(a) towards the center
(b) away from the center
(c) along a tangent
(d) will stop
Answer: (c)
When a particle moves in a circle with a speed v, its instantaneous velocity has a magnitude v and direction along the tangent at that point. The direction of the instantaneous velocity is kept changing by a centripetal force which is tension in the string in this case. As soon as the string breaks, this force disappears and the direction of the instant velocity of the particle can no longer change. So, it goes along the tangent to the instantaneous point on the circle.
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