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A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?
Answer:
w=50 rpm
Explanation:
Given data
The mass turntable M=2kg
Diameter of the turntable d=20 cm=0.2 m
Angular velocity ω=100 rpm= 100×(2π/60) =10.47 rad/s
Two blocks Mass m=500 g=0.5 kg
To find
Turntable angular velocity
Solution
We can find the angular velocity of the turntable as follow
Lets consider turntable to be disk shape and the blocks to be small as compared to turntable
[tex]I_{turntable}w=I_{block1}w^{i}+I_{turntable}w^{i}+I_{block2}w^{i}[/tex]
where I is moment of inertia
[tex]w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\ So\\I_{turntable}=M\frac{r^{2} }{2}\\I_{turntable}=2*(\frac{(0.2/2)}{2} )\\ I_{turntable}=0.01 \\And\\I_{block1}=I_{block2}=mr^{2}\\I_{block1}=I_{block2}=(0.5)*(0.2/2)^{2} \\ I_{block1}=I_{block2}==0.005\\so\\w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\w^{i}=\frac{0.01*(10.47)}{0.005+0.005+0.01} \\w^{i}=5.235 rad/s\\w^{i}=5.235*(60/2\pi )\\w^{i}=50 rpm[/tex]
The turntable's final angular velocity, just after the event is equal to 49.98 rpm.
Given the following data:
Mass of turntable = 2.0 kg.
Diameter = 20 cm to m = 0.2 m.
Mass of block = 500 g to kg = 0.5 kg.
Initial angular velocity = 100 rpm to rad/s = 10.47 rad/s.
First of all, we would determine the moment of inertia of the turntable and blocks respectively:
For the turntable:
[tex]I_T=\frac{1}{2} mr^2\\\\I_T=\frac{1}{2} \times 2 \times (0.1)^2\\\\I_T=0.01 \;kgm^2[/tex]
For the blocks:
[tex]I_B=\frac{1}{2}\times (2m)r^2\\\\I_B=0.5 \times (0.1)^2\\\\I_B=0.005 \;kgm^2[/tex]
Now, we would determine the turntable's final angular velocity by applying the law of conservation of momentum:
[tex]I_T\omega_f +I_{B1}\omega_f+I_{B2}\omega_f =I_T\omega_i\\\\\omega_f=\frac{I_T\omega_i }{I_T\omega_f +I_{B1}\omega_f+I_{B2}\omega_f } \\\\\omega_f=\frac{0.01 \times 10.47}{0.01 +0.05+0.05} \\\\\omega_f=\frac{0.1047}{0.11} \\\\\omega_f=5.235\;rad/s\\\\[/tex]
In RPM, we have:
[tex]\omega_f=5.235\times \frac{60}{2\pi} \\\\\omega_f= 49.98\;rpm[/tex]
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