Physics Topics cover a broad range of concepts that are essential to understanding the natural world.
Comparison Between Linear And Rotational Motion
In the discussion of rotational motion, some of the physical quantities are the rotational analogues of the corresponding physical quantities of linear motion. These quantities are given below:
Linear motion | Rotational motion |
Linear coordinate (x) | Angualr coordinate(θ) |
Displacement (s) | Angular displacement (θ) |
Velocity (v) | Angular velocity (ω) |
Acceleration (a) | Angular acceleration (α) |
Numerical Examples
Example 1.
A round table is rotated with an angular velocity of 10 rad ᐧ s-1 about its axis. Two blocks of mass m1 = 10 kg and m2 = 5 kg, connected to each other by a weightless inextensible string of length 0. 3 m, are placed along a diameter of the table. The coefficient of friction between the table and ml is 0. 5, while there is no friction between m2 and the table. Mass ml is at a distance of 0.124 m from the centre of the table. The masses are at rest with respect to the table,
(i) Calculate the frictional force on m1.
(ii) What should be the minimum angular speed of the table so that these masses will slip from the table?
(iii) How should these masses be kept so that the string remains taut but no frictional force acts on m1 ?
Solution:
The top view of the shown in Fig.
According to the problem,
AB = 0.3 m;
OA = 0.124 m
So, OB = 0.3 – 0.124
= 0.176 m
i) Centrifugal force acting on mass m1 along the direction OA,
m1ω2r = 10 × (10)2 × 0.124 = 124 N
Again, centrifugal force acting on mass m2 along the direction OB,
m2ω2r = 5 × (10)2 × 0.176 = 88 N
So, the resultant force along the direction OA = 124-88 = 36 N
In spite of this resultant force, the two masses remain stationary with respect to the table. Hence, the frictional force is equal and opposite to this force.
Hence, the required frictional force = 36 N.
ii) Reaction force of the table on mass m1 = weight of the mass = m1g = 10 × 9.8 = 98 N
As the coefficient of friction is 0.5, the limiting frictional force = 0.5 × 98 =49 N.
If the minimum angular velocity of the table is co, the resultant of the two opposite centrifugal forces acting on the two masses
= m1ω2r1 – m2ω2r2
= ω2(10 × 0.124 – 5 × 0.176)
= ω2(1.24 – 0.88) = 0.36ω2 N
According to the problem, resultant of the two centrifugal forces = limiting frictional force
∴ 0.36 ω2 = 49 or, ω2 = \(\frac{4900}{36}\)
or, ω = \(\frac{70}{6}\) = 11.67 rad ᐧ s-1
iii) If there is no frictional force acting on the masses, the resultant of the two centrifugal forces should be zero. In that case, if mass m1 is placed at a distance r from the centre, then the distance of m2 becomes (0.3 – r).
So, for any value of ω,
m1ω2r = m2ω2(0.3 – r) or, 0.3 – r = \(\frac{m_1}{m_2} r\)
or, 0.3 – r = 2r or, 3r = 0.3 or, r = 0.1 m
Example 2.
What should be the maximum speed of a motor car of mass 2000 kg when it takes a circular turn of radius 100 m on a plane road? Coefficient of friction between the tyre and the road = 0.25. Given, g = 10 m ᐧ s-2.
Solution:
Let the maximum speed of the motor car = v;
mass of the car = m,
radius of the circular path = r;
coefficient of friction between the tyre and the road = µ.
In the case of maximum speed of the car,
centripetal force = limiting frictional force
or, \(\frac{m v^2}{r}\) = µmg
or, v = \(\sqrt{\mu r g}\) = \(\sqrt{0.25 \times 100 \times 10}\) = 15.81 m ᐧ s-1
Example 3.
The roadway bridge over a stream is in the form of an arc of a circle of radius r. A car is crossing the bridge with a speed v. Prove that the limiting speed of the car with which it can cross the bridge without leaving the ground at the highest point of the bridge is v ≤ \(\sqrt{g r}\).
Solution:
Let the mass of the car be m and speed of the car be v.
∴ Weight of the car = mg and centrifugal force = \(\frac{m v^2}{r}\).
At the highest point on the bridge, the car will not jump up from the road if the weight of the car ≥ centrifugal force.
∴ mg ≥ \(\frac{m v^2}{r}\), gr ≥ v2
or, v2 ≤ gr, v ≤ \(\sqrt{g r}\)
Example 4.
What will be the angular velocity of the diurnal motion of the earth, so that the weight of a body at the equatorial region is 0.6 of the present weight? Radius of the earth = 6400 km.
Solution:
Due to the diurnal motion of the earth, a centrifu-gal force acts on a body on the surface of the earth and hence the body suffers an apparent loss of weight.
Apparent weight of the body at the equatorial region = mg – mω2R
(ω = angular velocity, R = radius of the earth)
According to the problem, 0.6 mg = mg – mω2R
or, 0.4g = ω2R
or, ω = \(\sqrt{\frac{0.4 \times g}{R}}\) = \(\sqrt{\frac{0.4 \times 9.8}{6400 \times 1000}}\) = 7.8 × 10-4 rad ᐧ s-1
Example 5.
A cylindrical drum made of steel and of diameter 20 cm is rotating about its own vertical axis. A small body made of steel remains stuck inside the cylinder when the drum rotates at the rate of 200 rpm. When the velocity of rotation decreases, the body falls down. What is the coefficient of friction between the body and the surface of the drum?
Solution:
Let the point where the body remains stuck be P [Fig.],
For the equilibrium of the body, weight of the body = limiting frictional force or, mg = F = µR
[R = normal force on the body by the wall of the drum]
Again, R supplies the necessary centripetal force to the body for its revolution.
Hence, R = \(\frac{m v^2}{r}\) = mω2r
∴ mg = µmω2r
or, µ = \(\frac{g}{\omega^2 r}\) = \(\frac{980 \times 9}{(20 \pi)^2 \times 10}\) [∵ ω = \(\frac{200 \times 2 \pi}{60}\) = \(\frac{20 \pi}{3}\)]
Example 6.
Four balls of mass 5 kg each are placed on the top of a horizontal turntable and fastened together with four strings of length 1 m each to form a square of side 1 m. The axis of rotation passes through the centre of the square. Find the tension in the strings when the turntable is rotated at the rate of \(\frac{30}{\pi}\) rpm.
Solution:
In the Fig. the table is shown from the top. When the turntable keeps on rotating, the centrifugal force experienced by each ball will be mω2r.
Here, m = 5 kg, ω = 2π × \(\frac{30}{\pi \times 60}\) = 1 rad ᐧ s-1
As the length of each side of the square is 1 m,
r = \(\frac{1}{2}\) × length of the diagonal or, r = \(\frac{\sqrt{2}}{2}\) = \(\frac{1}{\sqrt{2}}\) m
According to the figure, tension in each string,
T = mω2rcos45° = 5 × (1)2 × \(\frac{1}{\sqrt{2}}\) × \(\frac{1}{\sqrt{2}}\) = \(\frac{5}{2}\) = 2.5 N.
Example 7.
A hemispherical bowl of radius r is rotated with an angular velocity ω about a vertical axis passing through its centre. An object rotates with the same angular velocity remaining attached with the inner surface of the bowl. The straight line joining the body and the centre of the bowl makes an angle 45° with the vertical. The coefficient of friction between the body and the bowl is 0.2. Assuming the body is just about to move down along the curved surface of the bowl, prove that ω2r = 0.943 g.
Solution:
The hemisphere is rotated about the vertical axis AO with an angular velocity ω. The forces acting on the body of point P are shown in the Fig.
When the body is just about to move down,
mg = ncosθ + µnsinθ
or, mg = n(cosθ + µsinθ) ……. (1)
and mω2rsinθ = nsinθ – µncosθ
or, mω2r = n(1 – µcotθ) ….. (2)
From (1) and (2), we get
\(\frac{\omega^2 r}{g}\) = \(\frac{1-\mu \cot \theta}{\cos \theta+\mu \sin \theta}\)
= \(\frac{1-0.2 \cot 45^{\circ}}{\cos 45^{\circ}+0.2 \sin 45^{\circ}}\) [∵ µ = 0.2 and θ]
ω2r = 0.943g
Example 8.
A car of mass m is moving with a velocity v over a bridge. What will be the values of the force F at the highest point of a convex bridge and at the lowest point of a concave bridge?
Solution:
i) A centripetal force is required to move the car over a convex bridge. The value of this centripetal force is \(\frac{m v^2}{r}\)
Here, r is the radius of curvature of the convex bridge [Fig.],
At the highest point of the bridge, the resultant of the weight of the car mg and F will supply the necessary centripetal force.
∴ mg – F = \(\frac{m v^2}{r}\)
or, F = m(g – \(\frac{v^2}{r}\))
ii) A centripetal force is required to move the car over a concave bridge also. At the lowest point of the concave bridge, the resultant of the weight of the car mg and F will supply the necessary centripetal force [Fig.],
∴ F – mg = \(\frac{m v^2}{r}\) or, F = m(g + \(\frac{v^2}{r}\))
Example 9.
The roadway bridge over a stream is in the form of an arc of a circle of radius 50 m. what is the maximum speed with which a car can cross the bridge without leaving the ground at the highest point? [WBJEE’01]
Solution:
Suppose the car is moving with a speed v on the bridge so that it can cross the bridge without leaving the ground at the highest point [Fig.].
If the normal force at the highest point on the bridge is F, then
mg – F = \(\frac{m v^2}{r}\)
For the maximum speed of the car, at the highest point on the bridge, F = 0
∴ \(\frac{m v^2}{r}\) = mg or, v2 = rg
or, v = \(\sqrt{r g}\) = \(\sqrt{50 \times 9.8}\)
= 22 m ᐧ s-1
Example 10.
A car passes over a convex bridge. The centre of gravity of the car follows an arc of a circle of radius 30 m. Assuming that the car has a mass of 1000 kg, find the force that the car will exert at the highest point on the bridge, if the velocity of the car is 15 m ᐧ s-1. At what speed will the car lose contact with the road?
Solution:
Mass of the car, m = 1000 kg,
velocity, v = 15 m ᐧ s-1
and radius of the circular path, r = 30 m
The weight of the car = mg and the necessary centripetal force = \(\frac{m v^2}{r}\)
Let at the highest point on the convex bridge, the effective normal force by the road be F. Then
F = mg – \(\frac{m v^2}{r}\) = m(g – \(\frac{v^2}{r}\)) = 103(9.8 – \(\frac{15^2}{30}\))
= 2.3 × 103N
[∵ g = 9.8 m ᐧ s-1, v = 15 m ᐧ s-1 and r = 30 m]
When F = 0, the car loses contact with the road. In that situation, if the speed of the car is u, then
mg – 0 = \(\frac{m u^2}{r}\) or, u2 = gr
∴ u = \(\sqrt{9.8 \times 30}\) = 17.1 m ᐧ s-1.