Answer:
The correct solution is "64 RPM".
Explanation:
The given values are:
Mass,
M = 2.3 kg
Diameter,
D = 20 cm
i.e.,
= 0.2 m
Rotates at,
N = 110 rpm
Mass of block,
m = 460 g
i.e.,
= 0.46 kg
According to angular momentum's conservation,
⇒ [tex]I_1\omega_1=I_2\omega_2[/tex]
then,
⇒ [tex]I_1=\frac{1}{2}MR_2[/tex]
On substituting the values, we get
⇒ [tex]=\frac{1}{2}\times 2.3\times (0.1)^2[/tex]
⇒ [tex]=\frac{1}{2}\times 0.023[/tex]
⇒ [tex]=0.0115 \ kg \ m^2[/tex]
Now,
⇒ [tex]I_2=I_1+2mR^2[/tex]
[tex]=0.0115+2\times 0.46\times (0.1)^2[/tex]
[tex]=0.0115+0.0092[/tex]
[tex]=0.02 \ kg \ m^2[/tex]
then,
⇒ [tex]0.0115\times 110=0.02\omega_2[/tex]
⇒ [tex]1.265=0.02\omega_2[/tex]
⇒ [tex]\omega_2=\frac{1.265}{0.02}[/tex]
⇒ [tex]=63.25 \ or \ 64 \ RPM[/tex]
