Ask question

Ask question

All categories

3 years ago

12

A merry-go-round is shaped like a uniform disk and has moment of inertia of 50,000 kg m 2 . It is rotating so that it has an ang

ular momentum of 10,000 (kg m 2 radians/s) and its outer edge has a speed of 2 m/s. What is its radius, in m

1 answer:

lapo4ka [179]3 years ago

5 0

Answer:

r = 20 m

Explanation:

The formula for the angular momentum of a rotating body is given as:

L = mvr

where,

L = Angular Momentum = 10000 kgm²/s

m = mass

v = speed = 2 m/s

r = radius of merry-go-round

Therefore,

10000 kg.m²/s = mr(2 m/s)

m r = (10000 kg.m²/s)/(2 m/s)

m r = 5000 kg.m ------------- equation 1

Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:

I = (1/2) m r²

where,

I = moment of inertia = 50000 kg.m²

Therefore,

50000 kg.m² = (1/2)(m r)(r)

using equation 1, we get:

50000 kg.m² = (1/2)(5000 kg.m)(r)

(50000 kg.m²)/(2500 kg.m) = r

<u>r = 20 m</u>

You might be interested in

What equation p=m x v used to determine

OverLord2011 [107]

Answer:

The Momentum Calculator uses the formula p=mv, or momentum (p) is equal to mass (m) times velocity (v). The calculator can use any two of the values to calculate the third.

Explanation:

5 0

3 years ago

Two rooms are in contact. One is at 15°C and the other at 21°C. Choose the best statement concerning the heat flow.

amid [387]

A is the answer to your question

7 0

3 years ago

Read 2 more answers

A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds

aleksklad [387]

<em>There is not enough data to solve the problem, but I'm assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem</em>

Answer:

(a) $\Delta P=1.67 \ kg.m/s$

(b) $\Delta P=0.86\ m/s$

Explanation:

<u>Change of Momentum</u>

The momentum of a given particle of mass m traveling at a speed v is given by

$P=m.v$

When this particle changes its speed to a value v', the new momentum is

$P'=m.v'$

The change of momentum is

$\Delta P=m.v'-m.v$

$\Delta P=m.(v'-v)$

Defining upward as the positive direction, we'll compute the change of momentum in two separate cases.

(a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it's dropped to the floor:

$U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J$

This potential energy is transformed into kinetic energy just before the collision occurs, thus

$\displaystyle \frac{1}{2}mv^2=6.0368$

Solving for v

$\displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}$

$v=-14\ m/s$

This is the speed of the ball just before the collision with the floor. It's negative because it goes downward. Now we'll compute the speed it has after the collision. We'll use the new height and proceed similarly as above. The new height is

$h'=88.5\% (10)=8.85\ m$

The potential energy reached by the ball at its rebound is

$U'=m.g.h'=0.0616\cdot 9.8\cdot 8.85 =5.342568\ J$

Thus the speed after the collision is

$\displaystyle v'=\sqrt{\frac{5.342568\cdot 2}{0.0616}}$

$v'=13.17\ m/s$

The change of momentum is

$\Delta P=0.0616\cdot (13.17+14)$

$\Delta P=1.67 \ kg.m/s$

(b) If the putty sticks to the floor, then v'=0

$\Delta P=0.0616\cdot (0+14)$

$\Delta P=0.86\ m/s$

3 0

4 years ago

A cannon ball fired horizontally from a cliff has a velocity directed at 60 degrees above horizontal when it hits the ground 3.0

Kamila [148]

Since the ball is fired horizontally, the initial y velocity is zero and the time to hit the ground is the same as if the ball was simply dropped from the cliff. So you can solve the y position function:
$-\frac{1}{2}g t^{2} + h_{o}=0$
giving a height of 44.1m.
The given final velocity vector tells us that the initial x-directed velocity was about 17m/s.

3 0

3 years ago

Read 2 more answers

Some kids are playing by rolling a skateboard across level ground beside a 0.75 m high wall. The goal is to drop from the wall v

lys-0071 [83]

Answer:

The speed of the kid and the skateboard together is 0.723 m/s

Explanation:

Let the mass of the kid be $m_{k} = 45 kg$

Let the mass of the skateboard be $m_{s} = 4,5 kg$

The initial speed of the skateboard, $v_{s} = 8 m/s$

The initial horizontal speed of the boy as he jumps on the skateboard, $v_{k} = 0 m/s$

Let the speed of the kid and the skateboard together be v

According to the principle of conservation of momentum:

$m_{k} v_{k} + m_{s} v_{s} = (m_{k} + m_{s}) v$

Substituting the appropriate values into the given equation

$(45*0) + (4.5*8) = (45 + 4.8) v$

36 = 49.8 v

v = 36/49.8

v = 0.723 m/s

7 0

4 years ago