Question

A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round with a mass of 300 kg is spinning at 16 rpm . John runs tangent to the merry-go-round at 5.4 m/s , in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg. What is. What is the merry-go-round's angular velocity, in rpm, after John jumps on?

Answer 1

Answer:

$W_s = 19.05 rpm$    

Explanation:

For answer this we will use the law of the conservation of the angular momentum L where:

$L_i = L_f$

First, the important data is:

$M_t = 300 kg$  (mass of the merry-go-round)

$R_ t = 1.5 m$     (radius of the merry-go-round)

$W_t = 16 rpm = 1.675 rad/s$ (angular velocity of the merry-go-round)

$V_j =5.4m/s$ (velocity od jhon)

$M_j = 30 kg$  (mass of jhon)

Then,

$L_i = L_f$

$I_tW_t + M_jV_jR_t = I_sW_s$

where $I_s$ is the moment of inerta after jhon jumps on, $I_t$  is the moment of inertia of the merry-go-round and $W_s$ is the angular velocity of the merry-go-round after jhon jumps on.

So, we have to find the moment of inertia of the merry-go-round as:

$I_t = \frac{1}{2}M_tR_t^2$

$I_t = \frac{1}{2}(300)(1.5)^2$

$I_t = 337.5 kg*m^2$

Now we have to find$I_s$ as:

$I_s = I_t + M_jR_t^2$

$I_s = 337.5 + (30)(1.5)^2$

$I_s = 405 kg*m^2$

Then, we replace the data on the initial equation and we get:

$(337.5)(1.675) + (30)(5.4)(1.5) = (405)W_s$

Solving for $W_s$:

$W_s = 1.995 rad/s$

Finally, we change it to rpm as:

$W_s = 19.05 rpm$