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A CD has a playing time of 74.0 minutes. When the music starts, the CD is rotating at an angular speed of 480 revolutions per minute (rpm). At the end of the music, the CD is rotating at 210 rpm. Find:

(a) The magnitude of the average angular acceleration of the CD. Express your answer in rad/s2.
(b) The magnitude of the linear (or tangential) velocity at a point 3.00 cm from the center of rotation at the end of music. Express your answer in m/s.
(c) The radial acceleration at a point 3.00 cm from the center of rotation at the end of music. Express your answer in m/s2.
(d) The total distance a point 3.00 cm from the center of rotation passes during the play. Express your answer in m.

Respuesta :

Answer:

α =6.369*10⁻³ rad/s²

v₂= 0.6597 m/s

a₂= 14.5 m/s²

d = 4812.3 m

Explanation:

α =angular acceleration

ω = angular velocity

f = frequency

t = time

r= radius

v= linear velocity

d= distance

θ= angle covered for a distance d

a) the angular acceleration is the change in time of angular velocity, then

ω = 2π*f

α = (ω₂-ω₁)/ t = 2π*(f₂-f₁)/ t = 2π*(210 rpm - 480 rpm )/ 74 min = -22.925 rad/min²= 6.369*10⁻³ rad/s²

b) the linear velocity at the end of the music is

v₂= ω₂*r = 2π*f₂*r = 2π*210 rpm* 3 cm = 3958.4 cm/min=  0.6597 m/s

c) the radial acceleration at the end of the music is

a₂= ω₂²*r =  4π²*(210 rpm)²*3 cm = 5222994 cm/min² =  14.5 m/s²

d) the distance d covered under a constant angular acceleration is

ω₂²=ω₁²+2*α*θ

θ = (ω₂² - ω₁²)/(2*α) = 2π²(f₂² - f₁²)/α = 2π²[(210 rpm)² - (480 rpm)²)]/(-22.925 rad/min²) = 160410 rad

then the distance will be

d = r*θ = 3cm * 160410 rad = 481230 cm = 4812.3 m

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