# Physics Question HELPPP!!!?

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 5.0 revolutions.

If the CD has a radius of 6.5 cm and a mass of 20 g , what is the torque exerted on it?

### 3 Answers

- oubaasLv 72 months ago
angular speed ω = 2PI/60*n = 0.10472*450 = 47.12 rad/sec

acceleration angle Θ = 5*2PI = 31.42 radians

angular acceleration α = ω^2/2Θ = 47.12^2/62.84 = 35.34 rad/sec^2

M.o.I. J = m/2*r^2 = 10*10^-3*6.5^2*10^-4 = 4.225*10^-5 kg*m^2

torque T = J*α = 4.225*10^-5*35.34 = 1.5*10^-3 N*m (just 2 sign. digits)

- az_lenderLv 72 months ago
The rotational inertia of a disk spinning on its own axis is (1/2)MR^2, so in this case it's

(1/2)(0.020 kg)(0.065 m)^2 = 4.225 x 10^(-5) kg*m^2.

The final angular speed is (450 * 2pi rad)/(60 s) = 47.12 rad/s, so

the angular acceleration is (47.12 rad/s)^2 / (2*5*2pi rad) = 35.34 rad/s^2.

Torque = (I)(alpha)

= (4.225 x 10^(-5) kg*m^2)(35.34 s^(-2))

= 0.00149 N*m

- billrussell42Lv 72 months ago
450 rev/min x 1 min/60 s = 7.5 rev/s = ω

ω = Δθ/Δt

5 rev = 10π rad

Δt = Δθ/ω = 10π/7.5 = 4.189 s

α = Δω/Δt = 7.5/4.189 = 1.79 rad/s²

τ = αI

I = ½MR² = ½0.02(0.065)²

τ = 1.79(½)0.02(0.065)² = 0.0000756 Nm

I is moment of inertia in kg•m²

I = cMR²

M is mass (kg), R is radius (meters)

c = 1 for a ring or hollow cylinder

c = 2/5 solid sphere around a diameter

c = 7/5 solid sphere around a tangent

c = ⅔ hollow sphere around a diameter

c = ½ solid cylinder or disk around its center

c = 1/4 solid cylinder or disk around a diameter

c = 1/12 rod around its center, R = length

c = ⅓ for a rod around its end, R = length

c = 1 for a point mass M at a distance R from the axis of rotation

c = 1/3 for a door, where r is the width