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

$F_c = m\omega^2 R$

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

$F_c = 453.6 N$

Answer 2

1. The magnitude of the initial angular momentum of the merry-go-round is $285.18 \; Kgm^2/s^2$

2. The magnitude of the final angular momentum of the person 2 meters before she jumps on the merry-go-round is 663.07 $Kgm^2/s^2$.

3. The magnitude of the angular momentum of the person just before she jumps on to the merry-go-round is 663.07 $Kgm^2/s^2$.

4. The angular speed of the merry-go-round after the person jumps on is 2.04 rad/s.

5. The force needed by the person to hold on is 563.15 Newton.

Given the following data:

• Radius = 1.99 meters

• Moment of inertia I = 194 $Kgm^2$.

• Initial angular speed, ω = 1.47 rad/s

• Mass = 68 Kg

• Velocity = 4.9 m/s

1. To determine the magnitude of the initial angular momentum of the merry-go-round:

Mathematically, the initial angular momentum of an object is given by:

$L_i = I_i\omega_i\\\\L_i = 194 \times 1.47\\\\L_i = 285.18 \; Kgm^2/s^2$

2. To find the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round:

$L_f = mvR\\\\L_f = 68\times 4.9 \times 1.99$

Angular momentum, $L_f$ = 663.07 $Kgm^2/s^2$.

3. The magnitude of the angular momentum of the person just before she jumps on to the merry-go-round is the same as above because it is constant.

Angular momentum, L = 663.07 $Kgm^2/s^2$.

4. To find the angular speed of the merry-go-round after the person jumps on, we would use the law of conservation of angular momentum:

Applying the law of conservation of angular momentum:

Initial angular momentum + Final angular momentum = Initial angular momentum + Final angular momentum

$L_i + L_f = I_i\omega + I_f\omega\\\\L_i + L_f = (I_i + I_f)\omega\\\\L_i + L_f = (I_i + mR^2)\omega\\\\285.18 + 663.07 = (194 + 68[1.99]^2)\omega\\\\945.28 = (194 + 269.29)\omega\\\\945.28 = 463.29\omega\\\\\omega=\frac{945.28}{463.29} \\\\\omega= 2.04 \;rad/s$

5. To find the force needed by the person to hold on, we would use the centripetal force formula:

$F_c = m\omega^2 R\\\\F_c = 68 \times 2.04^2 \times 1.99\\\\F_c = 135.32 \times 4.1616\\\\F_c = 563.15 \; Newton$

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