Question

A 2.3 kg , 20-cm-diameter turntable rotates at 110 rpm on frictionless bearings. Two 460 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. Part A What is the turntable's angular velocity, in rpm, just after this event

Answer 1

The correct solution is "64 RPM".

Step-by-step 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,

⇒
`I_1\omega_1=I_2\omega_2`

then,

⇒
`I_1=(1)/(2)MR_2`

On substituting the values, we get

⇒
`=(1)/(2)* 2.3* (0.1)^2`

⇒
`=(1)/(2)* 0.023`

⇒
`=0.0115 \ kg \ m^2`

Now,

⇒
`I_2=I_1+2mR^2`

`=0.0115+2* 0.46* (0.1)^2`

`=0.0115+0.0092`

`=0.02 \ kg \ m^2`

then,

⇒
`0.0115* 110=0.02\omega_2`

⇒
`1.265=0.02\omega_2`

⇒
`\omega_2=(1.265)/(0.02)`

⇒
`=63.25 \ or \ 64 \ RPM`