Laws of Motion
NCERT Solution
Part 4
Question 31: A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/g. The mass of the train is 106 kg. What provides the centripetal force required for this purpose: the engine or the rails? What is the angle of banking required to prevent wearing out of the rail?
Answer: Given, radius r = 30 m, speed of train v = 54 km/h = `54xx5/(18)=15` m/s, mass m = 106 kg
Rail is providing the centripetal force.
Angle of banking can be calculated as follows:
Tan θ = `(v^2)/(rg)`
`=(15^2)/(30xx10)=(225)/(300)=0.75`
Or, θ = 36.87°
Question 32: A block of mass 25 kg is raised by a 50 kg man in two different ways. In first case, the man is pulling up the block of mass with a rope. In second case, the man is using a pulley to pull up the mask whereby he pulls down the rope over pulley to lift the mass up. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?
Answer: Mass of block mb = 25 kg, mass of man mm = 50 kg, g = 10 m s-2
Force applied on block = mg = 25 × 10 = 250 N
Force applied on man = mg = 50 × 10 = 500 N
When the man is directly pulling the block he is applying force upwards. It increases his apparent weight. In this case, F = 500 + 250 = 750 N
When the man is pulling the block through pully, he is applying a downward force. It decreases his apparent weight. In this case, F = 500 – 250 = 250 N
As the floor yields to force of 700 N, he should prefer the second mode.
Question 33: A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey
(a) Climbs up with an acceleration of 6 m s-2
Answer: m = 40 kg, a = 6 m s-2, Tmax = 600 N
Tension in rope can be calculated as follows:
T – mg = ma
Or, T = m(a + g)
= 40(6 + 10) = 640 N
Here, T > Tmax, so rope will break.
(b) Climbs down with an acceleration of 4 m s-2
Answer: mg – T = ma
Or, T = m(g – a)
= 40(10 – 6) = 160 N
Here, T < Tmax, so rope will not break.
(c) Climbs up with a uniform speed of 4 m s-1
Answer: Uniform speed means acceleration is zero
T = m(a + g) = 40 × 10 = 400 N
Here, T < Tmax, so rope will not break.
(d) Falls down the rope nearly freely under gravity?
Answer: When the monkey falls freely under gravity, its acceleration becomes almost equal to g
So, mg – T = ma
Or, T = m(g – g) = 0
Here, T < Tmax, so rope will not break
Question 34: Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall. The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between μs and μk.
Answer: Mass of A = mA = 5 kg, mass of B = mB = 10 kg, F = 200 N, μs = 0.15
Force of friction can be calculated as follows:
`f_s=μ(m_A+m_B)g`
= 0.15 (5 + 10) × 10 = 22.5 N (Towards left)
Net force acting on partition = 200 – 22.5 = 177.5 N (Towards right)
Reaction force of partition = 177.5 N (Towards left, as per Third Law of Motion)
Force of friction on mass A can be calculated as follows:
`f_A=μm_A\g`
= 0.15 × 5 × 10 = 7.5 N (Towards left)
Net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N (Towards right)
Mass be exerts an equal amount of force on mass A towards left.
When the wall is removed, the two bodies will move towards right, i.e. in the direction of applied force.
In this case, net force acting on the moving system = 177.5 N
Now acceleration of the system can be calculated as follows:
F = (mA + mB)a
Or, `a=F/(m_A+m_B)`
`=(177.5)/(5+10)=(177.5)/(15)=11.83` m s-2
Now, net force causing mass A to move
FA = mAa
= 5 × 11.83 = 59.15 N
So, net force exerted by mass A on mass B = 192.5 – 59.15 = 133.35 N
This force will act in the direction of motion. An equal amount of force will exerted by mass B on A but in opposite direction, i.e. in opposite direction to motion.
Question 35: A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.
Answer: Mass of block m = 15 kg, μ = 0.18, a = 0.5 m s-1
Force on the block caused by motion of the trolley can be calculated as follows:
F = ma = 15 × 0.5 = 7.5 N (In the direction of motion of trolley)
Force of static friction between the block and the trolley can be calculated as follows:
`f=μmg`
= 0.18 × 15 × 10 = 27 N
Here, the force of static friction is greater than applied force. So, the block will appear at rest with reference to trolley.
For a stationary observer, the trolley will appear to be moving. For a moving observer the trolley will appear to be at rest, assuming his relative speed with reference to trolley is zero.
Question 36: The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end. The coefficient of friction between the box and the surface below is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).
Answer: Given, m = 40 kg, s = 5 m, μ = 0.15, acceleration of truck = 2 m s-2, u = 0, v = ?
Force on box F = ma = 40 × = 80 N
Frictional force Ff = μmg
= 0.15 × 40 × 10 = 60 N
So, net force = F – Ff
= 80 – 60 = 20 N
Backward acceleration produced in box = a = Net force ÷ m
`=(20)/(40)=0.5` m s-2
The box is placed 5 m away from rear end so it needs to cross 5 m in order to fall off. The time taken for this can be calculated as follows:
`s=ut+1/2at^2`
Or, `5=1/2xx0.5t^2`
Or, `t^2=(5xx2)/(0.5)=20`
Or, `t=sqrt(20)=2sqrt5` s
The distance covered by truck can be calculated as follows:
`s=ut+1/at^2`
`=1/2xx2xx(2sqrt5)^2`
`=4xx5=20` m
Question 37: A disc revolves with a speed of `33\1/2` min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?
Answer: Given, r = 15 cm = 0.15 m, frequency of revolutions `=(100)/3` rev/min = `5/9` rev/s, μ = 0.15
In this situation, if the coin has frictional force greater than or equal to the centripetal force it will revolve with the disc.
Coin at 4 cm:
Radius of revolution = r' = 4 cm = 0.04 m
Angular frequency = ω = 2πv
`=2xx3.14xx5/9=3.49` s-1
Frictional force f = μmg
`=0.15xxmxx10=1.5` m N
Centripetal force on coin
F = mr'ω2
`=m\xx0.04xx(3.49)^2=0.49` m N
As frictional force on this coin is greater than centripetal force, the coin will revolve along with the record.
Coin at 14 cm:
Radius of revolution r"e; = 14 cm = 0.14 m
Angular frequency = 3.49 s-1
Frictional force on coin = 1.5 m N
Centripetal force on coin
F = mr"e;ω2
`=mxx0.14xx(3.49)^2`
= 1.7 m N
As frictional force on coin is less than centripetal force so this coin will slip from the surface of the record.
Question 38: You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?
Answer: When a motrocylcist reaches the top of the loop, centripetal force is balanced together by normal reaction FN and weight (force due to gravity) Fg of motorcyclist. As a result, the motorcyclist does not fall.
Equation for centripetal acceleration can be written as follows:
Fnet = mac
Or, FN + Fg = mac
Or, FN + mg `=(mv^2)/r`
Or, `v^2=(mg\r)/m=gr`
Or, `v=sqrt(gr)`
We know that normal reaction is provided by speed of motorcycle. At minimum speed vmin, FN = 0
So, vmin `=sqrt(gr)`
`=sqrt(10xx25)=5sqrt(10)` m/s
Question 39: A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the maximum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
Answer: Given, m = 70 kg, r = 3 m, μ = 0.15, frequency of rotation = 200 rev/min `=(10)/3` rev/s
The required centripetal force for rotation of the man is provided by normal force FN.
When the floor revolves, the man sticks to the wall of the drum. So, weight of the man = mg (acting downward) is balanced by frictional force (f = μFN) acting downward.
So, the man will not fall till his weight is less than frictional force
Or, mg < f
Or, mg < μFN = μmrω2
Or, g < μrω2
Or, `ω=sqrt(g/(μr))`
`=sqrt((10)/(0.15xx3))`
`=2/3sqrt5` rad s-1
This is the minimum angular speed
Question 40: A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for ω ≤ `sqrt(g/R)`. What is the angle made by the radius vector joining the center to the bead with the vertical downward direction for ω = `sqrt((2g)/R)`? Neglect friction.
Answer: Let us assume that the radius vector joining the bead with the centre makes an angle θ with the vertical downward direction.
Radius R = OP and N = Normal reaction
Vertical and horizontal equations of forces can be written as follows:
mg = N cos θ ……………(1)
`ml(ω^2)` = N sin θ ……………(2)
In ΔOPQ
Sin θ = `l/R`
Or, `l=` R sin θ ………….(3)
Substituting this value in equation (2), we get
m(R sin θ) ω2 = N sin θ
Or, mR ω2 = N ……………….(4)
Substituting this in equation (1) we get
mg = mR ω2 cos θ
Or, cos θ = `g/(Rω^2)` ……………..(5)
As cos θ < 1, the bead will remain at its lowermost point
For `g/(Rω^2)` ≤ 1
Or, ω ≤ `sqrt(g/R)`
So, ω `=sqrt((2g)/R)`
Or, ω2 `=(2g)/R` ………….(6)
Equating equations (5) and (6)
`(2g)/R=g/(R co\sθ)`
Or, cos θ `=1/2`
Or, θ = 60°