Question

A merry-go-round with a a radius of R = 1.99 m and moment of inertia I = 194 kg-m2 is spinning with an initial angular speed of ω = 1.47 rad/s in the counter clockwise direection when viewed from above. A person with mass m = 68 kg and velocity v = 4.9 m/s runs on a path tangent to the merry-go-round. Once at the merry-go-round the person jumps on and holds on to the rim of the merry-go-round.1) What is the magnitude of the initial angular momentum of the merry-go-round? kg-m²/s2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round? kg-m²/s3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round? kg-m²/s4) What is the angular speed of the merry-go-round after the person jumps on? rad/s5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on? N

Answer 1

Answer:

Part 1)

$L_1 = 185.2 kg m^2/s^2$

Part 2)

$L_2 = 663.07 kg m^2/s^2$

Part 3)

$L = 663.07 kg m^2/s^2$

Part 4)

$\omega = 1.83 rad/s$

Part 5)

$F_c = 453.6 N$

Explanation:

Part a)

Initial angular momentum of the merry go round is given as

$L_1 = I \omega$

here we know that

$I = 194 kg m^2$

$\omega = 1.47 rad/s^2$

now we have

$L_1 = 194 \times 1.47$

$L_1 = 185.2 kg m^2/s^2$

Part b)

Angular momentum of the person is given as

$L = mvR$

so we have

$m = 68 kg$

$v = 4.9 m/s$

R = 1.99 m

so we have

$L_2 = (68)(4.9)(1.99)$

$L_2 = 663.07 kg m^2/s^2$

Part 3)

Angular momentum of the person is always constant with respect to the axis of disc

so it is given as

$L = 663.07 kg m^2/s^2$

Part 4)

By angular momentum conservation of the system we will have

$L_1 + L_2 = (I_1 + I_2)\omega$

$185.2 + 663.07 = (194 + 68(1.99^2))\omega$

$848.27 = 463.28 \omega$

$\omega = 1.83 rad/s$

Part 5)

Force required to hold the person is centripetal force which act towards the center

so we will have

$F_c = m\omega^2 R$

$F_c = 68(1.83^2)(1.99)$

$F_c = 453.6 N$