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• CBSE Class 11 Physics Notes

# Motion in a Vertical Circle

When analyzing the motion of a body in a vertical circle, we must take gravity into account. The magnitudes of the body’s velocity and tension in the string fluctuate continually due to the impact of the earth’s gravitational field. At the lowest position, it is maximum, and at the highest position, it is minimal. As a result, the circular motion in a vertical circle is not uniform.

### Expression for Velocity of Body Moving in a Vertical Circle

Consider a tiny body with mass ‘m’ that is swirled in a vertical circle with radius ‘t’ by one end of a string. The body’s acceleration rises as it descends the vertical circle and decreases as it ascends the vertical circle in this example. As a result, the body’s speed is constantly changing.

It reaches its maximum at the bottom of the vertical circle and its minimum at the top. As a result, the body does not move circularly. The weight ‘mg’ always operates vertically downward, regardless of where the particle is on the circle. Motion in a Vertical Circle

Let ‘L’ represent the vertical circle’s lowest point.

Let ‘u’ represent the body’s velocity at L.

Let ‘v’ represent the body’s velocity at any point P on a vertical circle.

Let ‘h’ represent the distance between point P and point L.

By the law of conservation of energy,

The energy at point P = Energy at point L

(½) m v2 + m g h = (½) m u2

v2 + 2 g h = u2

v2 = u2 – 2 g h

v = √(u2 – 2 g h)

This is an equation for a particle’s velocity at any point in a vertical circle while it is moving in a circular motion.

### Expression for Tension in String in Motion of Body in Vertical Circle

Consider the centripetal force at point P.

T – m g cosθ = m v2 ⁄ r

T = m g cosθ + m v2 ⁄ r

From figure, cosθ = (r – h) ⁄ r

Substitute in the equation of T.

T = m g (r – h) ⁄ r + m v2 ⁄ r

= m ⁄ r (g (r – h) + v2)

But v2 = u2 – 2 g h

T = m ⁄ r (g r – g h + u2 – 2 g h)

= m ⁄ r (u2 – 3 g h + g r)

This is an equation for tension in string.

### Special Cases for Tension

Case 1: When the body is at lowermost position (h = 0)

We have,

T = m ⁄ r (u2 – 3 g h + g r)

At point L (h = 0)

TL = m ⁄ r (u2 – 3 g (0) + g r)

= m ⁄ r (u2 + g r)

Case 2: When the body is at uppermost position (h = 2 r)

We have,

T = m ⁄ r (u2 – 3 g h + g r)

At point H (h = 2 r)

TH = m ⁄ r (u2 – 3 g (2 r) + g r)

= m ⁄ r (u2 – 5 g r)

Case 3: When the string is horizontal (h = r)

T = m ⁄ r (u2 – 3 g h + g r)

At point M (h = r)

TM = m ⁄ r (u2 – 3 g (r) + g r)

= m ⁄ r (u2 – 2 g r)

Relationship between tension at the highest point and at the lowest point

TL – TH = m ⁄ r (u2 + g r) – m ⁄ r (u2 – 5 g r)

= m u2 ⁄ r + m g – m u2 ⁄ r + 5 m g

= 6 m g

As a result, the tension in the string at the lowest point L is six times larger than the tension at the highest position H.

### Minimum Velocity of Body at Different Positions When Looping a Loop

(1) At Lowest Point L (h = 0)

This is the lowest body velocity necessary for the body to circle a loop, or travel around the circle once entirely. Therefore, at the highest point, tension must be greater than 0.

TH > 0

m ⁄ r (u2 – 5 g r) > 0

u2 – 5 g r > 0

u2 > 5 g r

u > √(5 g r)

This is the required minimum velocity at the lowest point of the vertical circle.

(2) At Highest Point H (h = 2r)

We have,

v = √(u2 – 2 g h)

= √(5 g r – 2 g (2 r))

= √(5 g r – 4 g r)

= √(g r)

(3) When String is Horizontal (h = r)

We have,

v = √(u2 – 2 g h)

= √(5 g r – 2 g r)

= √(3 g r)

This is the required minimum velocity when the string is horizontal.

• If the velocity of the body is such that v < √(2 g r), then the body oscillates about the lowest point of the vertical circle.
• If the velocity of the body is such that √(2 g r) < v < √(5 g r), then the body leaves the circular path and acts as a projectile. It will leave circular motion between horizontal and highest point.

### Energy of Body Moving in a Vertical Circle

The energy of the body has two parts: Kinetic Energy and Potential Energy.

(1) At Lowest Point L (h = 0)

Kinetic Energy, EK = (1⁄2) m v2

= (1⁄2) m (√(5 g r))2

= (5⁄2) m g r

Potential Energy, EP = m g h

= m g (0)

= 0

Total Energy, ET = EK + EP

= (5⁄2) m g r + 0

= (5⁄2) m g r

(2) At Highest Point H (h = 2r)

• Kinetic Energy, EK = (1⁄2) m v2

= (1⁄2) m (√(g r))2

= (1⁄2) m g r

• Potential Energy, EP = m g h

= m g (2r)

= 2 m g r

• Total Energy, ET = EK + EP

= (1⁄2) m g r + 2 m g r

= (5⁄2) m g r

(3) When the string is horizontal (h = r)

• Kinetic Energy, EK = (1⁄2) m v2

= (1⁄2) m (√(3 g r))2

= (3⁄2) m g r

• Potential Energy, EP = m g h

= m g (r)

= m g r

• Total Energy, ET = EK + EP

= (3⁄2) m g r + m g r

= (5⁄2) m g r

The lowest point on the circular route has the most kinetic energy, whereas the highest point has the least. The total amount of energy is preserved.

### Sample Problems

Problem 1: A stone weighing 2 kg is whirled in a vertical circle at the end of a rope of length 1 m. Find the velocity of a stone at:

a) the lowest position

b) midway when the string is horizontal

c) topmost position to just complete the circle.

Solution:

Given:

Radius of circular path, r = 1 m

Mass of the body, m = 2 kg

Gravity, g = 9.8 m ⁄ s2

a) Velocity at lowermost point, vL = √(5 g r)

= √(5 × 9.8 × 1) m ⁄ s

= 7 m ⁄ s

b) When string is horizontal, vM = √(3 g r)

= √(3 × 9.8 × 1) m ⁄ s

= 5.42 m ⁄ s

c) Velocity at topmost point, vH = √(g r)

= √(9.8 × 1) m ⁄ s

= 3.13 m ⁄ s

Hence, velocity at lowermost point is 7 m ⁄ s, when string is horizontal is 5.42 m ⁄ s, and velocity at topmost point is 3.13 m ⁄ s.

Problem 2: A stone weighing 2 kg is whirled in a vertical circle at the end of a rope of length 1 m. Find the tension in the string at:

a) the lowest position

b) midway when the string is horizontal

c) topmost position to just complete the circle.

Solution:

Given:

Radius of circular path, r = 1 m

Mass of the body, m = 2 kg

Gravity, g = 9.8 m ⁄ s2

a) Tension at lowermost point, TL = m ⁄ r (u2 + g r)

= m ⁄ r (5 g r + g r)

= 6 m g

= 6 × 2 × 9.8 N

= 117.6 N

b) When string is horizontal, TM = m ⁄ r (u2 – 2 g r)

= m ⁄ r (5 g r – 2 g r)

= 3 m g

= 3 × 2 × 9.8 N

= 58.8 N

c) Tension at topmost point, TH = m ⁄ r (u2 – 5 g r)

= m ⁄ r (5 g r – 5 g r)

= 0 N

Hence, tension at lowermost point is 117.6 N, when string is horizontal is 58.8 N, and tension at topmost point is 0 N.

Problem 3: An object of mass 2 kg attached to a string of length 1 m is whirled in a vertical circle at constant angular speed if the maximum tension in the string is 10 kg wt. Calculate the speed of the object.

Solution:

Given:

Mass of object, m = 2 kg,

Radius of circle, r = 1 m,

Tension in the string, T = 10 kg wt

= 10 × 9.8 N

= 98 N

Tension in a vertical circle is maximum at lowest point.

Tmax = m v2 ⁄ r + m g

= m (v2 ⁄ r + g)

v2 ⁄ r + g = Tmax ⁄ m

v = √(r (Tmax ⁄ m – g))

= √(1 (98 ⁄ 2 – 9.8)) m ⁄ s

= √39.2 m ⁄ s

= 6.26 m ⁄ s

Hence, speed of the object is 6.26 m ⁄ s.

Problem 4: Find out the point in a path at which a particle may have zero tension in a vertical circle of radius ‘r’.

Solution:

Let the tension in the string at the highest point be T.

Minimum speed required by the particle at the highest point to complete the vertical circular motion is √ (g r).

Equation of equilibrium

m v2 ⁄ r = T + m g

m (g r) ⁄ r = T + m g

m g = T + m g

T = 0

Hence, tension can be zero at the highest point.

Problem 5: A car moves on a flat circular track of 100 m radius with a constant speed of 40 m ⁄ s. If the car weighs 500 kg, find out the centripetal force acting on the car.

Solution:

Given:

Mass of the car, m = 500 kg

Radius of the track, r = 100 m

Speed of the car, v = 40 m ⁄ s

Angular velocity, ω = v ⁄ r

= 40 ⁄ 100