Question

Each atv needs 4 wheels and each mini tricycle needs 3?

Answer 1

When three children jump on a merry-go-round, the moment of inertia increases by 25%, according to the conservation of angular momentum, the new rotation rate becomes 3.2 revolutions per minute.

Step-by-step explanation:

Merry-go-round Conservation of Angular Momentum

When the three children jump onto the spinning merry-go-round, they cause an increase in the moment of inertia of the system. According to the conservation of angular momentum, the product of the moment of inertia (I) and the angular velocity (ω) must remain constant if no external torques are applied. The problem states that the moment of inertia increases by 25%, so we can express this mathematically as:

1. Iinitial × ωinitial = Ifinal × ωfinal
2. Ifinal = Iinitial × 1.25
3. ωfinal = ωinitial / 1.25

Given that the initial rotation rate is 4.0 rev/min, the new rotation rate will be:

• 4.0 rev/min / 1.25 = 3.2 rev/min

The new rotation rate of the merry-go-round after the children jump on is 3.2 revolutions per minute.

Answer 2

The new rotation rate of the merry-go-round after the children jump on, causing a 25% increase in the moment of inertia, is calculated using the conservation of angular momentum and is found to be 3.2 rev/min.

Step-by-step explanation:

The question addresses the concept of angular momentum and its conservation in the context of a merry-go-round. The conservation of angular momentum dictates that the product of the moment of inertia and the rotational speed (angular velocity) is constant for a closed system with no external torques.

Given that a merry-go-round is initially rotating at 4.0 rev/min and its moment of inertia increases by 25% due to three children jumping on it, we can determine the new rotation rate using the following steps:

Let I represent the initial moment of inertia of the merry-go-round and ω be the initial rotation rate (4.0 rev/min).

The conservation of angular momentum before and after the children jump on can be written as

I * ω = (I + 0.25I) * ω’, where ω’ is the new rotation rate.

Solving for ω’, we find that ω’ = (I * ω) / (1.25I) = ω/1.25.

Plugging in the initial rotation rate (4.0 rev/min), we get ω’ = 4.0 rev/min / 1.25 = 3.2 rev/min.

The new rotation rate of the merry-go-round after the children jump on is 3.2 rev/min.