Suppose you observe your friend (mass 66.6 kg) stand at the very edge of a motionless merry-go-round (modelled as a thin disk with moment of inertia 5.7 kg m 2). At some point, your friend walks around and along the edge at a speed 1.4 m/s perfectly tangent to the edge at all times, relative to the merry-go-round’s surface. Take the radius of the merry-go-round to be 1.3 m.

(a)
Calculate the speed (in m/s) of your friend relative to the stationary ground.

(b)
Calculate the speed (in m/s) of your friend relative to the stationary ground if the merry-go-round becomes massless.
(c)
Calculate the speed (in m/s) of your friend relative to the stationary ground if the merry-go-round becomes arbitrarily (infinitely) massive.

To solve this problem, we can use the principle of conservation of angular momentum. The initial angular momentum of your friend and the merry-go-round system is equal to the final angular momentum.

Given:
Mass of the friend (m1) = 66.6 kg
Moment of inertia of the merry-go-round (I) = 5.7 kg m^2
Radius of the merry-go-round (r) = 1.3 m
Speed of the friend relative to the merry-go-round (v) = 1.4 m/s

(a) To find the speed of your friend relative to the stationary ground:

The initial angular momentum (L_initial) of the system is given by:
L_initial = m1 * v1 * r

Since the merry-go-round is motionless initially, the initial angular momentum is zero.

The final angular momentum (L_final) of the system is given by:
L_final = m1 * v_final * r + I * ω

Where,
v_final is the speed of your friend relative to the stationary ground.
ω is the angular velocity of the merry-go-round.

Since your friend walks along the edge of the merry-go-round, the angular velocity (ω) of the merry-go-round is equal to v_final / r.

Setting the initial and final angular momenta equal to each other, we have:
0 = m1 * v_final * r + I * (v_final / r)

Simplifying,
0 = m1 * v_final * r + (I/r) * v_final

0 = (m1 * r + I/r) * v_final

Solving for v_final,
v_final = 0 / (m1 * r + I/r)

Since the denominator is zero,
v_final = 0 m/s

Therefore, the speed of your friend relative to the stationary ground is 0 m/s when the merry-go-round becomes motionless.

(b) To find the speed of your friend relative to the stationary ground if the merry-go-round becomes massless:

When the merry-go-round becomes massless, the moment of inertia (I) becomes zero.

Using the equation from part (a):
v_final = 0 / (m1 * r + I/r)

With I = 0,
v_final = 0 / (m1 * r + 0/r)

v_final = 0 / (m1 * r)

v_final = 0 m/s

Therefore, the speed of your friend relative to the stationary ground is still 0 m/s when the merry-go-round becomes massless.

(c) To find the speed of your friend relative to the stationary ground if the merry-go-round becomes arbitrarily (infinitely) massive:

When the merry-go-round becomes infinitely massive, the moment of inertia (I) becomes very large.

Using the equation from part (a):
v_final = 0 / (m1 * r + I/r)

As I becomes very large, the second term in the denominator dominates, and we can neglect the first term.

Therefore,
v_final ≈ 0 / (I/r)

Since I/r gives the angular velocity at the edge of the merry-go-round (ω), we can rewrite the equation as:
v_final ≈ 0 / ω

Since ω is the angular velocity, which is constant for an infinitely massive merry-go-round,
v_final ≈ 0 / ω = 0 m/s

Therefore, the speed of your friend relative to the stationary ground is 0 m/s when the merry-go-round becomes infinitely massive.

To solve this problem, we can apply the principle of conservation of angular momentum. Angular momentum is the product of moment of inertia and angular velocity. According to the conservation of angular momentum, the initial angular momentum of the system should be equal to the final angular momentum.

Let's start by calculating the initial angular momentum of the system. Since your friend is standing at the very edge of the motionless merry-go-round, his initial angular momentum is zero (as he is not moving or rotating).

(a) To calculate the speed of your friend relative to the stationary ground, we need to find the final angular velocity of the merry-go-round when your friend is walking around along its edge. We can use the equation:

Initial Angular Momentum = Final Angular Momentum

Since the initial angular momentum is zero, we have:

0 = (moment of inertia of the merry-go-round) * (final angular velocity of the merry-go-round)

The moment of inertia of a thin disk is given by: I = 0.5 * mass * radius^2

Substituting the values given in the question: mass of the merry-go-round = 66.6 kg and radius = 1.3 m, we can calculate the moment of inertia of the merry-go-round:

I = 0.5 * 66.6 kg * (1.3 m)^2
I = 56.67 kg*m^2

Rearranging the equation, we find:

(final angular velocity of the merry-go-round) = 0 / 56.67 kg*m^2
(final angular velocity of the merry-go-round) = 0 rad/s

Since the final angular velocity is zero, the speed of your friend relative to the stationary ground is also zero.

(b) Now, let's consider the scenario where the merry-go-round becomes massless. In this case, the moment of inertia of the merry-go-round would be zero.

Using the equation from above:

0 = (moment of inertia of the merry-go-round) * (final angular velocity of the merry-go-round)

Substituting moment of inertia as zero:

0 = 0 * (final angular velocity of the merry-go-round)

Since any number multiplied by zero is zero, the final angular velocity of the merry-go-round could be any value. Therefore, the speed of your friend relative to the stationary ground would be determined by the final angular velocity, which is not specified in the question.

(c) Lastly, let's consider the scenario where the merry-go-round becomes arbitrarily massive, which means its moment of inertia becomes infinitely large.

Using the equation from the first part:

0 = (moment of inertia of the merry-go-round) * (final angular velocity of the merry-go-round)

Since the moment of inertia is infinitely large, even if the final angular velocity is non-zero, the final angular velocity multiplied by infinity would still be zero. Therefore, the speed of your friend relative to the stationary ground would again be zero.

123

sdfgsdg