Question

A phonograph turntable initially rotating at 3.5 rad/s makes 3 complete turns before coming to a stop. what is its angular acceleration?​

Answer 1

Explanation:

a) The angle in radians that corresponds to 3 rev is

q = (3 rev)(2p rad/rev) = 6p rad

From the formula, , we finda

w w

q p = − = − ( )

( )( ) = − f o

2 2 2

2

0 3 5

2 6

0 325 . . rad/s

rad

rad/s

2

Answer 2

The angular acceleration of the phonograph turntable is -0.32 rad/s².

Given:

Initial angular velocity= ω₀= 3.5 rad/s

Final angular velocity= ω= 0 ( As the phonograph turntable stopped after some time)

Number of revolutions= 3

To find:

The angular acceleration of the phonograph turntable.

Solution:

• The rate of change of angular velocity with respect to time of an object is known as angular acceleration.

Since, 1 revolution= 2π rad

So, 3 revolution= 3×2π rad

θ= Angular distance = 3×2×π rad

We can find the angular acceleration using the formula,

ω²=ω₀²+2αθ ( where ω= Final angular velocity, ω₀= Initial angular velocity, α= angular acceleration and θ= angular displacement)

Put the values of ω, ω₀ and θ in the above formula.

ω²=ω₀²+2αθ

0²=(3.5)²+2×α×2×3×π

0=12.25+12απ

-12.25=12απ

α=$\frac{-12.25}{12\pi }$

Put the value of π as 3.14.

α= -0.32 rad/s²

Therefore, the angular acceleration of the phonograph turntable is -0.32 rad/s².

# SPJ1