Answer:
15.99861 seconds
1865.58658 Joules
Explanation:
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation
I = Moment of inertia of disc = 0.031 kgm²
t = Time taken
Initial angular speed
[tex]\omega_i=3313\frac{2\pi}{60}\\\Rightarrow \omega_i=346.93\ rad/s[/tex]
Final angular velocity
[tex]\omega_f=0.75\times \omega_i\\\Rightarrow \omega_f=0.75\times 346.93\\\Rightarrow 260.19\ rad/s[/tex]
From the equation of rotational motion
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{260.19-346.93}{4}\\\Rightarrow \alpha=-21.685\ rad/s^2[/tex]
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-346.93}{-21.685}\\\Rightarrow t=15.99861\ s[/tex]
Time taken by the turntable to slow down is 15.99861 seconds
Rotational energy is given by
[tex]K=\frac{1}{2}I\omega_i^2\\\Rightarrow K=\frac{1}{2}\times 0.031\times 346.93^2\\\Rightarrow K=1865.58658\ J[/tex]
The work has to be done on the turntable to bring it to rest is 1865.58658 Joules
The amount of time it would take the record player turntable to come to rest is 15.99 seconds and the amount of work done is 1865.59 Joules.
First of all, we would determine the final angular velocity of this record player turntable as follows:
ωf = 75% × ωi
Note: ωi = 3313 × 2π/60 = 346.93 rad/s.
ωf = 0.75 × 346.93
ωf = 260.19 rad/s.
Next, we would determine the angular acceleration by using the first equation of rotational motion:
ωf = ωi + αt
α = (ωf - ωi)/t
α = (260.19 - 346.93)/4
α = -21.685 rad/s².
Similarly, the time taken is given by:
t = (ωf - ωi)/α
t = (0 - 346.93)/-21.685
Time, t = 15.99 seconds.
Mathematically, the rotational energy possessed by this record player turntable is equal to the work done and it can be calculated by using this formula:
K.E = 1/2 × Iωi²
K.E = 1/2 × 0.031 × 346.93²
K.E = 1865.59 Joules.
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