Question

The magnitude of displacement of a particle moving in a circle of radius 5 m with angular displacement of 60° with respect to initial position is​

Answer 1

Answer:

Tangential acceleration, a

T

=r

dt

dω

Given, θ=t

2

−2t+7

Angular velocity, ω=

dt

dθ

=2t−2

Now,

dt

dω

=2

Thus, a

T

=20×2=40cm/s

2

=0.4m/s

2

Answer 2

Given :

• Particle is moving in a circle of radius, r = 5 m
• Angular displacement of Particle, θ = 60°

To find :

• Magnitude of linear Displacement of particle, s = ?

Formula required :

• Relation between Linear and angular displacement during the uniform circular motion

     

       θ = r / s

[ where θ is angular displacement in Radian, r is radius of circle in metres, s is linear displacement in metres]

Solution :

Converting angular displacement given in degrees into Radian unit

→ θ = 60°

→ θ = 60 × π / 180

→ θ = π / 3  Rad

Calculating linear displacement of particle

→ θ = r / s

→ π / 3 = 5 / ( s )

→ s = 5 × 3 / π

→ s = 15 / π   m

putting value of π

→ s = 15 / ( 22 / 7 )

→ s = 15 × 7 / 22

→ s = 4.773   m   [ approx. ]

therefore,

• Magnitude of Linear displacement of particle is 4.773 metres. (approximately).