A particle describes uniform circular motion in a circle of radius 2 m, with the angular speed of 2 rad s-1 . The magnitude of the change in its velocity in (π/2) s is

Question

A particle describes uniform circular motion in a circle of radius 2 m, with the angular speed of 2 rad s-1 . The magnitude of the change in its velocity in (π/2) s is (A) 0 ms-1 (B) 2√2 ms-1 (C) 8 ms-1 (D) 4 ms-1

Tardigrade Admin · Accepted Answer

Given, r=2 m , ω=2 rad s -1, t=(π/2) S Now, θ=ω t=π rad V =r ω=2 × 2 V =4 ms -1 Now, Δ v=2 v sin (θ/2) Δ v =2 × 4 × sin (π/2) Δ v =8 ms -1

Q. A particle describes uniform circular motion in a circle of radius $2 m$ , with the angular speed of $2 r a d s^{- 1}$ . The magnitude of the change in its velocity in $\frac{π}{2} s$ is

$0 m s^{- 1}$

$22 m s^{- 1}$

$8 m s^{- 1}$

$4 m s^{- 1}$

$42 m s^{- 1}$

Given, $r = 2 m, ω = 2 r a d s^{- 1}, t = \frac{π}{2} S$
Now, $θ = ω t = π r a d$
$V = r ω = 2 \times 2$
$V = 4 m s^{- 1}$
Now, $Δ v = 2 v sin \frac{θ}{2}$
$Δ v = 2 \times 4 \times sin \frac{π}{2}$
$Δ v = 8 m s^{- 1}$

Q. A particle describes uniform circular motion in a circle of radius 2m, with the angular speed of 2rads−1 . The magnitude of the change in its velocity in 2π​s is

0ms−1

22​ms−1

8ms−1

4ms−1

42​ms−1

Given, r=2m,ω=2rads−1,t=2π​S Now, θ=ωt=πrad V=rω=2×2 V=4ms−1 Now, Δv=2vsin2θ​ Δv=2×4×sin2π​ Δv=8ms−1