Questions OBJECTIVE - I and Answer (Circular Motion) HC Verma Part 1 (9-16)
Q#9
A coin placed on a rotating turntable just slips if it is placed at a distance of 4 cm from the center. If the angular velocity of the turntable is doubled it will just slip at a distance of
(a) 1 cm
(b) 2 cm
(c) 4 cm
(d) 8 cm
Answer: (a)
The friction force to keep it rotating = mω²r. At 4 cm away from the center when it just slips, the static frictional force is at its limiting value. When the angular speed of the turntable is doubled the magnitude of the required centripetal force increases four times.
So, to keep it to limiting value of friction force, 'r' should be made one fourth. It will be clear as below
m(2ω)².r/4 = mω²r.
That is why it should be placed at 1 cm.
(a) increases
(b) decreases
(c) remains the same
(d) fluctuates
Answer: (a)
An over-bridge is not just an arch shape placed on a horizontal road because it will give a great jerk at the start. So to keep the driving smooth both ends of the over-bridge are kept concave upward. When a vehicle starts to ascend on this concave upward part of the over-bridge it puts extra force on the road in addition to the weight due to movement on the circular part. So the apparent weight and hence the normal force on the vehicle increases.
Q#11
Three identical cars A, B and C are moving at the same speed on three bridges. The car A goes on a plane bridge, B on a bridge convex upward and C goes on a bridge concave upward. Let FA, FBand FC be the normal forces exerted by the cars on the bridges when they are at the middle of the bridges.
(a) FA is maximum of the three forces.
(b) FB is maximum of the three forces.
(c) FC is maximum of the three forces.
(d) FA = FB = FC
Answer: (c)
Normal force on a plane bridge is equal to weight of the car. On a convex upward bridge due to the upward inertial force apparent weight reduces and so is the normal force on the vehicle. But at the middle of concave upward bridge this inertial force is downward making the apparent weight greater. Therefore the normal force is maximum in this case.
Q#12
A train 'A' runs from east to west and another train 'B' of the same mass runs from west to east at the same speed along the equator. 'A' presses the track with a force F1and 'B' presses the track with a force F2 .
(a) F1 > F2
(b) F1 < F2
(c) F1 = F2
(d) The information is insufficient to find the relation between F1 and F2.
Answer: (a)
The earth spins from west to east at equator. To maintain the same speed the first train has to press the track more than normal as it goes against the speed of earth's surface while the second train gets help from the earth's movement and it has to press the track less than the normal which explains the answer.
Q#13
If the earth stops rotating, the apparent value of g on its surface will
(a) increases everywhere
(b) decreases everywhere
(c) remains the same everywhere
(d) increases at some places and remain the same at some places.
Answer: (d)
If the earth stops rotating the apparent weight will increase at most of the places because the upward component of the centrifugal force due to rotation will disappear. But at poles it will not change because these places do not rotate in a circle on the earth's surface.
Q#14
A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. Let T1and T2 be the tensions at the points L/4 and 3L/4 away from the pivoted ends.
(a) T1 > T2
(b) T2 > T1
(c) T1 = T2
(d) The relation between T1 = T2 depends on whether the rod rotates clockwise or anticlockwise.
Answer: (a)
Taking the rod of uniform mass per unit length = m. Mass beyond a distance r from the pivot = m(L – r). And this mass will be assumed to be concentrated at middle of the (L – r) length ie at CG. So, distance of the CG from the pivot
= r + (L - r)/2 = (L + r)/2
Force due to circular motion on this length = Tension at point r away from the pivot
= m(L – r)ω²(L + r)/2
= mω²(L² - r²)/2
It is clear from this expression that as r increases Tension in the rod decreases. It explains the answer. See the figure below. (Please Note: Instead of asking this question as Objective Type or Multiple choice, it can be asked as "Prove that/Show that T1 > T2. In that case you can prove it with diagram as shown above.)
Q#15
A simple pendulum having a bob of mass 'm' is suspended from the ceiling of a car used in a stunt film shooting. The car moves up along an inclined cliff at a speed v and makes a jump to leave the cliff and lands at some distance. Let R be the maximum height of the car from the top of the cliff. The tension in the string when the car is in air is
(a) mg
(b) mg – mv²/r
(c) mg + mv²/r
(d) zero
Answer: (d)
During the jump of the car from the cliff its trajectory is like a projectile and from the maximum height it is freely falling under gravity with an acceleration g downward. Taking it as frame of reference, it is a non-inertial frame of reference. So a pseudo force equal to mg has to be applied to the bob in upward direction which neutralizes the weight and making it zero. So tension in the string is zero.
Q#16
Let θ denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the bob is 'm', the tension in the string is mg.cosθ
(a) always
(b) never
(c) at the extreme positions
(d) at the mean position.
Answer: (c)
At any position tension in the string is = mgcosθ + mv²/r
(Where r is length of pendulum and v is instantaneous velocity of bob)
v is zero only at extreme positions. So, tension is mg.cosθ only at extreme positions.
(a) 1 cm
(b) 2 cm
(c) 4 cm
(d) 8 cm
Answer: (a)
The friction force to keep it rotating = mω²r. At 4 cm away from the center when it just slips, the static frictional force is at its limiting value. When the angular speed of the turntable is doubled the magnitude of the required centripetal force increases four times.
So, to keep it to limiting value of friction force, 'r' should be made one fourth. It will be clear as below
m(2ω)².r/4 = mω²r.
That is why it should be placed at 1 cm.
Q#10
A motorcycle is going on an over-bridge of radius R. The driver maintains a constant speed. As the motor cycle is ascending on the over-bridge, the normal force on it(a) increases
(b) decreases
(c) remains the same
(d) fluctuates
Answer: (a)
An over-bridge is not just an arch shape placed on a horizontal road because it will give a great jerk at the start. So to keep the driving smooth both ends of the over-bridge are kept concave upward. When a vehicle starts to ascend on this concave upward part of the over-bridge it puts extra force on the road in addition to the weight due to movement on the circular part. So the apparent weight and hence the normal force on the vehicle increases.
Q#11
Three identical cars A, B and C are moving at the same speed on three bridges. The car A goes on a plane bridge, B on a bridge convex upward and C goes on a bridge concave upward. Let FA, FBand FC be the normal forces exerted by the cars on the bridges when they are at the middle of the bridges.
(a) FA is maximum of the three forces.
(b) FB is maximum of the three forces.
(c) FC is maximum of the three forces.
(d) FA = FB = FC
Answer: (c)
Normal force on a plane bridge is equal to weight of the car. On a convex upward bridge due to the upward inertial force apparent weight reduces and so is the normal force on the vehicle. But at the middle of concave upward bridge this inertial force is downward making the apparent weight greater. Therefore the normal force is maximum in this case.
Q#12
A train 'A' runs from east to west and another train 'B' of the same mass runs from west to east at the same speed along the equator. 'A' presses the track with a force F1and 'B' presses the track with a force F2 .
(a) F1 > F2
(b) F1 < F2
(c) F1 = F2
(d) The information is insufficient to find the relation between F1 and F2.
Answer: (a)
The earth spins from west to east at equator. To maintain the same speed the first train has to press the track more than normal as it goes against the speed of earth's surface while the second train gets help from the earth's movement and it has to press the track less than the normal which explains the answer.
Q#13
If the earth stops rotating, the apparent value of g on its surface will
(a) increases everywhere
(b) decreases everywhere
(c) remains the same everywhere
(d) increases at some places and remain the same at some places.
Answer: (d)
If the earth stops rotating the apparent weight will increase at most of the places because the upward component of the centrifugal force due to rotation will disappear. But at poles it will not change because these places do not rotate in a circle on the earth's surface.
Q#14
A rod of length L is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. Let T1and T2 be the tensions at the points L/4 and 3L/4 away from the pivoted ends.
(a) T1 > T2
(b) T2 > T1
(c) T1 = T2
(d) The relation between T1 = T2 depends on whether the rod rotates clockwise or anticlockwise.
Answer: (a)
Taking the rod of uniform mass per unit length = m. Mass beyond a distance r from the pivot = m(L – r). And this mass will be assumed to be concentrated at middle of the (L – r) length ie at CG. So, distance of the CG from the pivot
= r + (L - r)/2 = (L + r)/2
Force due to circular motion on this length = Tension at point r away from the pivot
= m(L – r)ω²(L + r)/2
= mω²(L² - r²)/2
It is clear from this expression that as r increases Tension in the rod decreases. It explains the answer. See the figure below. (Please Note: Instead of asking this question as Objective Type or Multiple choice, it can be asked as "Prove that/Show that T1 > T2. In that case you can prove it with diagram as shown above.)
Q#15
A simple pendulum having a bob of mass 'm' is suspended from the ceiling of a car used in a stunt film shooting. The car moves up along an inclined cliff at a speed v and makes a jump to leave the cliff and lands at some distance. Let R be the maximum height of the car from the top of the cliff. The tension in the string when the car is in air is
(a) mg
(b) mg – mv²/r
(c) mg + mv²/r
(d) zero
Answer: (d)
During the jump of the car from the cliff its trajectory is like a projectile and from the maximum height it is freely falling under gravity with an acceleration g downward. Taking it as frame of reference, it is a non-inertial frame of reference. So a pseudo force equal to mg has to be applied to the bob in upward direction which neutralizes the weight and making it zero. So tension in the string is zero.
Q#16
Let θ denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the bob is 'm', the tension in the string is mg.cosθ
(a) always
(b) never
(c) at the extreme positions
(d) at the mean position.
Answer: (c)
At any position tension in the string is = mgcosθ + mv²/r
(Where r is length of pendulum and v is instantaneous velocity of bob)
v is zero only at extreme positions. So, tension is mg.cosθ only at extreme positions.
 HC Verma Part 1 (9-16)){kind=link}
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