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

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,

⇒  $I_1\omega_1=I_2\omega_2$

then,

⇒  $I_1=\frac{1}{2}MR_2$

On substituting the values, we get

⇒      $=\frac{1}{2}\times 2.3\times (0.1)^2$

⇒      $=\frac{1}{2}\times 0.023$

⇒      $=0.0115 \ kg \ m^2$

Now,

⇒  $I_2=I_1+2mR^2$

        $=0.0115+2\times 0.46\times (0.1)^2$

        $=0.0115+0.0092$

        $=0.02 \ kg \ m^2$

then,

⇒  $0.0115\times 110=0.02\omega_2$

⇒              $1.265=0.02\omega_2$

⇒                  $\omega_2=\frac{1.265}{0.02}$

⇒                       $=63.25 \ or \ 64 \ RPM$