Question

towards left from the vertical.
13. A particle of mass 5 g moves in a circle of radius 25 cm at the rate of 2 rev/second. Find the
acceleration of the particle and centripetal force acting on it. (UPB 2006) Ans. 39.4 ms
ITOI
the vertical ; (iv) 0.02 cm/s" downwards at
0.197 N.
A body of 1 kg mass, tied with a thread of length 10 m​

Answer 1

Question :

A particle of mass 5 g moves in a circle of radius 25 cm at the rate of 2 revolutions/second. Find the acceleration of the particle and centripetal force acting on it.

Given :

• Mass of the Particle = 5 g
• Radius of the Circle = 25 cm
• No. of revolution per Seconds = 2 r/s.
• Time Taken for 2 revolutions = 1 s

To find :

• The Angular Acceleration of the Particle.
• Centripetal force acting on the Particle.

Solution :

Here, Given radius is 25 cm.

So let's convert it in m.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀1 m = 100 cm

⠀⠀⠀⠀⠀⠀⠀⠀⠀25 cm = 25/100 m

⠀⠀⠀⠀⠀⠀⠀25 cm = 0.25 m

So, first let us find the circumference of the Circle :

We know the Formula for circumference of a circle ,i.e,

$\boxed{\bf{C = 2\pi r}}$

Where :

• C = Circumference of the Circle
• r = Radius of the Circle.

So using the formula for circumference of a circle and substituting the values in it,we get :

$:\implies \bf{C = 2\pi r}$

$:\implies \bf{C = 2 \times \dfrac{22}{7} \times 0.25}$

$:\implies \bf{C = 2 \times \dfrac{22}{7} \times 0.25}$

$:\implies \bf{C = \dfrac{11}{7}}$

$:\implies \bf{C = 1.571\:(Approx.)}$

$\boxed{\therefore \bf{C = 1.571\:m}}$

Hence the circumference of the Circle is 1.571 m.

Angular velocity of the particle :

We know that :

$\boxed{\bf{\omega = \dfrac{\Delta\:s}{\Delta\:t}}}$

Where :

• ω = Angular velocity
• ∆ s = Angle of Rotation
• ∆ t = Time

Here :

• ∆ s = 1.561
• ∆ t = 1 s

$:\implies \bf{\omega = \dfrac{\Delta\:s}{\Delta\:t}}$

$:\implies \bf{\omega = \dfrac{2 \times 1.571}{1}}$

[Note : Here, the ∆ s is multiplied by 2 as the revolution is 2 revolutions / second.]

$:\implies \bf{\omega = 2 \times 1.571}$

$:\implies \bf{\omega = 3.142}$

$\boxed{\therefore \bf{\omega = 3.142\:ms^{-1}}}$

Hence the angular velocity of the Particle is 3.142 m/s.

To find the Angular acceleration :

We know the formula for Angular acceleration i.e,

$\boxed{\bf{a = \dfrac{v^{2}}{r}}}$

Where :

• a = Angular acceleration
• v = Angular velocity
• r = Radius of the Circle

Using the formula for Angular acceleration and substituting the values in it, we get :

$:\implies \bf{a = \dfrac{v^{2}}{r}}$

$:\implies \bf{a = \dfrac{(3.142)^{2}}{0.25}}$

$:\implies \bf{a = \dfrac{9.872}{0.25}}$

$:\implies \bf{a = 39.5}$

$\boxed{\therefore \bf{a = 39.5\:ms^{-2}}}$

Hence, the Angular Acceleration of the particle is 39.5 m/s².

To find the Centripetal force Acting on the Particle :

Here, Given mass is 5 g.

So let's convert it in kg.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀1 g = 1000 kg

⠀⠀⠀⠀ 5 g = 5/1000 kg

⠀⠀⠀⠀⠀⠀⠀5 g = 0.005 kg

We know the formula for centripetal force i.e,

$\boxed{\bf{F_{c} = \dfrac{mv^{2}}{r}}}$

Where :

• F = Centripetal force
• m = Mass of the particle
• r = Radius of the Circle

Using the formula for Centripetal force and substituting the values in it, we get :

$:\implies \bf{F_{c} = \dfrac{mv^{2}}{r}}$

$:\implies \bf{F_{c} = \dfrac{0.005 \times 3.142^{2}}{0.25}}$

$:\implies \bf{F_{c} = \dfrac{0.005 \times 9.872}{0.25}}$

$:\implies \bf{F_{c} = \dfrac{0.049}{0.25}}$

$:\implies \bf{F_{c} = 0.1975(approx.)}$

$\boxed{\therefore \bf{F_{c} = 0.2}}$

Hence, the centripetal force is 0.2 N