Question

A mass of 2 kg is moving along a circular path of radius 1 m. If the mass moves with 300 revolutions per minute, its kinetic energy would be

Answer 1

Answer:

The answer is 100π²

Explanation:

• KE = 1/2 mv²
• Here m is given and also r is also given as 1m
• Also, it's given that it makes 300 rev/min, ie, 300/60 rps.
• ∴  For 1rev = 2π rad, for 5 rps, 2π × 5 = 10 rad/s, ie, ω = 10π rad/s
• ∵ v = Rω, v = 1m × 10πrad/s = 10π rad/s²
• ∴ KE = 1/2 × 2kg × (10π rad/s²)²
• ∴ 100 π² rad/s²

Hoped it helped...... :)

Answer 2

Answer:

The kinetic energy of the mass is 986.96J

Relations to be noted:

                                     ω = 2π * No . of revolutions

                                     v = rω

where v is the velocity, r is the radius of rotation, ω is the angular velocity.

Explanation:

Given the mass of the body = 2 kg

  radius of the circular path = 1 m.

   No. of revolutions = 300 rpm = 5 rps

Now, Angular velocity ω = 2π*5 = 10π rad/s.

   velocity v = rω = 10π rad/s

Kinetic energy = $\frac{mv^{2} }{2} = \frac{2*(10\pi)^{2} }{2} = 100\pi ^{2} = 986.96 J$