Solved Examples Work, Energy & Power - Physics Class 11 - NEET

Some solved Examples 

Ex.1 Two blocks each of mass M are connected to the ends of a light frame as shown in figure. The frame is rotated about the vertical line of symmetry. The rod breaks if the tension in it exceeds T₀. Find the maximum frequency with which the frame may be rotated without breaking the rod.  

Solution: Consider one of the blocks at a distance l from the axis of rotation.
Angular speed is ω = 2 p f.
Centripetal acceleration of the block is ω² l.
Centripetal force required is M ω² l, which is supplied by the tension in the rod.
At the breaking limit the tension equals T₀; therefore
T₀ = M ω² l
Thus the maximum angular speed is
ω_max = √(T₀ / (M l))
Converting to frequency,
f_max = ω_max / (2 p) = (1 / (2 p)) √(T₀ / (M l))

Ex.2 Prove that a motor car moving over a convex bridge is lighter than the same car resting on the same bridge.  

Solution: The car moves over the convex portion AB which may be approximated by a circular arc of radius R.
At the top of the arc the forces acting on the car of mass m are its weight mg (downwards) and the normal reaction N (upwards from the bridge).
These two forces provide the required centripetal force towards the centre of the circular arc.
Centripetal force required = m v² / R.
Net downward force = weight - normal reaction = mg - N.
Equating centripetal force to net downward force,
mg - N = m v² / R
Therefore
N = mg - m v² / R
Since m v² / R is positive for non-zero speed, R < mg.
Hence the normal reaction (the apparent weight) of the moving car is less than the weight of the stationary car; i.e., the car is lighter when moving over the convex bridge.

Ex.3 A body weighing 0.4 kg is whirled in a vertical circle with a string making 2 revolutions per second. If the radius of the circle is 1.2m. Find the tension (a) at the top of the circle, (b) at the bottom of the circle. Give : g = 10 m s^-2 and p = 3.14 

Solution: Mass of the body is m = 0.4 kg.
Frequency of revolution is f = 2 s^-1.
Angular speed is
ω = 2 p f
ω = 2 × p × 2 = 4 p
Numerically, using p = 3.14,
ω = 4 × 3.14 = 12.56 rad s^-1
Radius is r = 1.2 m.
Centripetal acceleration magnitude is
r ω² = 1.2 × (12.56)²
Evaluating,
r ω² ≈ 1.2 × 157.7536 ≈ 189.3043 m s^-2
Centripetal force magnitude = m r ω²
m r ω² ≈ 0.4 × 189.3043 ≈ 75.7217 N

(a) At the top of the circle the forces acting along the string (towards centre) are the tension T (upwards along the string towards centre) and the weight mg (also towards the centre when at the top). The required centripetal force is provided by their sum; thus
T + mg = m r ω²
Therefore
T = m r ω² - m g
Using g = 10 m s^-2,
T = 75.7217 - 0.4 × 10 = 75.7217 - 4 = 71.7217 N
T ≈ 71.7 N (at the top)

(b) At the bottom of the circle, the weight acts downwards (away from centre) while tension acts upwards (towards centre). The required centripetal force is provided by the difference between tension and weight; hence
T - m g = m r ω²
Therefore
T = m r ω² + m g
T = 75.7217 + 4 = 79.7217 N
T ≈ 79.7 N (at the bottom)

Ex.4 A metal ring of mass m and radius R is placed on a smooth horizontal table and is set rotating about its own axis in such a way that each part of ring moves with velocity v. Find the tension in the ring.

Solution: Consider a small element ACB of the ring which subtends a small angle Δθ at the centre.
Let the tension in the ring be T. The two neighbouring parts of the ring pull the element ACB with tensions of magnitude T directed tangentially at A and B.
Resultant radial component of these two tensions directed towards the centre is
2 T sin(Δθ / 2)
For small Δθ, sin(Δθ / 2) ≈ Δθ / 2, hence resultant radial component ≈ T Δθ.
Length of the element ACB is R Δθ.
Linear mass density of the ring is λ = m / (2 p R).
Mass of the element is
Δm = λ × (R Δθ) = (m / (2 p R)) × (R Δθ) = (m / (2 p)) Δθ
This element moves in a circular path of radius R with speed v, so the required centripetal force on this element is
ΔF = Δm × v² / R = (m / (2 p)) Δθ × v² / R
Equating the radial component of tension to required centripetal force,
T Δθ = (m v² / (2 p R)) Δθ
Cancel Δθ from both sides to obtain
T = m v² / (2 p R)
This is the required tension in the rotating ring.

Ex.5 A small smooth ring of mass m is threaded on a light inextensible string of length 8L which has its ends fixed at points in the same vertical line at a distance 4L apart. The ring describes horizontal circles at constant speed with both parts of the string taut and with the lower portion of the string horizontal. Find the speed of the ring and the tension in the string. The ring is then tied at the midpoint of the string and made to perform horizontal circles at constant speed of . Find the tension in each part of the string. 

Solution: When the ring slides on the string with both parts taut, the magnitude of tension is the same in both parts.

From the geometry of the arrangement the distances from the lower pivot to the ring and from the upper pivot to the ring are given by

BP = 3L and AP = 5L

Let the angle each part of the string makes with the vertical be θ. The vertical component of tension in each part must support the weight of the ring.
Thus 2 T cos θ = m g
If the lower portion of the string is horizontal, the lower segment contributes no vertical component at that instant; the geometry given above yields one part shorter than the other and the appropriate trigonometric relations lead to
T cos θ = m g   ...(i)
The horizontal component of tension provides the centripetal force for circular motion. If the radius of the horizontal circle described by the ring is r, then
T sin θ = m v² / r   ...(ii)
Using the geometry of the triangle formed by the string segments and the vertical separation one may express r and sin θ, cos θ in terms of L, and then solve (i) and (ii) to find v and T.
In the second case, when the ring is tied at the midpoint the triangle ABP becomes equilateral and the tensions in the two equal halves need not be equal to the first case.
For the tied midpoint case the geometry gives an equilateral triangle ABP so the angles and side relations are fixed.
Using vertical and horizontal component balances and centripetal requirement yields the two tensions for the two parts; the algebraic solution follows from solving the corresponding component equations.

Ex.6 A large mass M and a small mass m hang at the two ends of the string that passes through a smooth tube as shown in Figure. The mass m moves around in a circular path, which lies in the horizontal plane. The length of the string from the mass m to the top of the tube is l and q is the angle this length makes with vertical. What should be the frequency of rotation of mass m so that M remains stationary ? 

Solution:

For the mass M to remain stationary, the tension in the string must balance its weight.

Consider the forces acting on mass m. The tension T in the string acts along the string making an angle θ with the vertical. The weight mg acts vertically downward.

Vertical equilibrium of mass m

Since the circular motion lies in a horizontal plane, the vertical acceleration of mass m is zero.
T cosθ = mg   ...(2)
Substituting T = Mg from equation (1) into equation (2):
Mg cosθ = mg
cosθ = m / M
This condition shows that the motion is possible only if M ≥ m.

Horizontal motion of mass m
The horizontal component of tension provides the centripetal force required for circular motion.
Radius of the circular path: r = l sinθ

Centripetal force equation: T sinθ = mω²r
T sinθ = mω²l sinθ
Cancelling sinθ (since θ ≠ 0):
T = mω²l   ...(3)
Using equation (1):
Mg = mω²l
ω = √(Mg / ml)
The frequency f is related to angular speed by: f = ω / (2π)
f = (1 / 2π) √(Mg / ml)

Final Answer

The required frequency of rotation of mass m so that mass M remains stationary is:
f = (1 / 2π) √(Mg / ml)
Condition: M ≥ m