Question

Angular acceleration of partical in circular motion which slows down 600 RPM to rest in 10seconds

Answer 1

Explanation:

α=ωf –ωi =0– 600 x 2π

t. 60

10

= –20π = –2π

10

Answer 2

Answer:

The angular acceleration of the particle in circular motion is -10rad/s².

Explanation:

We will use the following equation for solving this question:

$\omega=\omega_0+\alpha t$     (1)

Where,

ω=final angular velocity

ω₀=initial angular velocity

α=angular acceleration

t=time during which the motion occurs

From the question we have,

The initial angular velocity(ω₀)=600RMP=100RSP

The time during which the motion occurs(t)=10 seconds

The final angular velocity(ω)=0

Inserting the entities in equation (1) we get;

$0=100+\alpha \times 10$

$\alpha=\frac{-100}{10}$

$\alpha=-10 \ rad/s^2$

The angular acceleration of the particle in circular motion is -10rad/s².

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