Question

nly the final value of the answer. A merry-go-round with a moment of inertia of 6.78 x 109 kg • m2 is coasting at 2.20 rad/s. When a 87-kg man steps onto the rim, the angular velocity decreases to 20 rad/s. What is the radius in meters of the merry-go-round? Write your answer with ONE decimal place.

Answer 1

given
I_merrygoround = 6.78*10^3 kg/m^2
wi = 2.20 rad/s
m = 87 kg
wf = 2.0 rad/s

radius of the merry-go-round, R = ?

Apply conservation of angular momentum

If*wf = Ii*wi

(I_merrygoround + m*R^2)*wf = I_merrygoround*wi

(6.78*10^3 + 87*R^2)*2 = 6.78*10^3*2.2

(6.78*10^3 + 87*R^2) = 6.78*10^3*2.2/2

87*R^2 = 6.78*10^3*2.2/2 - 6.78*10^3

R^2 = (6.78*10^3*2.2/2 - 6.78*10^3)/87

R^2 = 7.793

R = sqrt(7.793)

= 2.8 m <<<<<<<<<<<---------------Answer

Answer 2

SOLUTION :

Let r meters be the radius of the merry-go-round.

Moment of inertia of the man weighing 87 kg

= 87 * r^2.       (Where r is the radius of the merry-go-round  ground)

M.I. of the merry-go-round is given = 6.78*10^3       kg-m^2

So, M.I. of after the man steps onto the merry-go-round

= 87 r^2 + 6.78*10^3     kg-m^2

Angular momentum remains unchanged .

So,

M.I. before l* angular speed before = M.I. afterwards * angular speed afterwards

=> 6.78*10^3 * 2.2 = (6.78*10^3 + 87 r^2) * 2.0

=> r^2 = (6.78*10^3 * 2.2/2.0 - 6.78*10^3) / 87

=> r^2 = 7.7931

=> r = 2.7916 m = 2.8 m approximately (ANSWER)