- #1

Freye

- 28

- 0

**1. A 4.5m diameter merry-go-round is rotating freely with an angular velocity of 0.70 rad/s; its total moment of inertia is 1750 kg X m/s^2. Four people standing on the ground each of 65 kg mass suddenly step onto the edge of the merry-go-round.**

a) What will be the angular velocity of the merry-go-round now? b)What if the people were on it initially, and then jumped off?

r= d/2 = 4.5/2 = 2.25m

w[tex]_{}1[/tex]= 0.70 rad/s

I[tex]_{}1[/tex]= 1750kg X m^2

I[tex]_{}people[/tex] = 65 X 4 X 2.25^2 = 1316.25kg X m^2

I[tex]_{}2[/tex]= I[tex]_{}1[/tex] + I[tex]_{}people[/tex] = 3066.25 Kg X m^2

a) What will be the angular velocity of the merry-go-round now? b)What if the people were on it initially, and then jumped off?

r= d/2 = 4.5/2 = 2.25m

w[tex]_{}1[/tex]= 0.70 rad/s

I[tex]_{}1[/tex]= 1750kg X m^2

I[tex]_{}people[/tex] = 65 X 4 X 2.25^2 = 1316.25kg X m^2

I[tex]_{}2[/tex]= I[tex]_{}1[/tex] + I[tex]_{}people[/tex] = 3066.25 Kg X m^2

**2. I[tex]_{}1[/tex]w[tex]_{}1[/tex] = I[tex]_{}2[/tex]w[tex]_{}2[/tex]**

T=rFsin[tex]\theta[/tex]

T=I[tex]\alpha[/tex]

T=rFsin[tex]\theta[/tex]

T=I[tex]\alpha[/tex]

**3. The correct answers as they are given in the back of the book are a) w[tex]_{}2[/tex]=0.40rad/s and b)no change (if jump off radially)**

a)

w[tex]_{2}[/tex] = I[tex]_{}1[/tex]w[tex]_{}1[/tex]/I[tex]_{}2[/tex]

= (1750)(.07)/(3066.25)

= 0.40 rad/s [the correct answer]

b)

w[tex]_{}1[/tex]= 0.40 rad/s [from part a)]

I[tex]_{}1[/tex] and I[tex]_{}2[/tex] from part a) switch places

w[tex]_{}2[/tex]=I[tex]_{}1[/tex]w[tex]_{}1[/tex]/I[tex]_{}2[/tex]

w[tex]_{}2[/tex] = (3066.25)(0.40)/(1750)

w[tex]_{}2[/tex]= 0.70 rad/s [which is to be expected, however apparently incorrect]

At this point, I realized that for some reason the conservation of angular momentum must not apply in this scenario, and so I did

[tex]\theta[/tex] = 0degrees (the angle between the radius and the applied force, assuming they jump of radially)

T=rfsin[tex]\theta[/tex]=0

[tex]\alpha[/tex]=T/I

[tex]\alpha[/tex]=0

Therefore [tex]\Delta[/tex]w=0 [the correct answer]

Essentially, my question is why the law of conservation of angular momentum applies when people step onto the merry-go-round, but not when they step off? I know I am making some sort of obvious logical blunder, but I cannot see it :(.

a)

w[tex]_{2}[/tex] = I[tex]_{}1[/tex]w[tex]_{}1[/tex]/I[tex]_{}2[/tex]

= (1750)(.07)/(3066.25)

= 0.40 rad/s [the correct answer]

b)

w[tex]_{}1[/tex]= 0.40 rad/s [from part a)]

I[tex]_{}1[/tex] and I[tex]_{}2[/tex] from part a) switch places

w[tex]_{}2[/tex]=I[tex]_{}1[/tex]w[tex]_{}1[/tex]/I[tex]_{}2[/tex]

w[tex]_{}2[/tex] = (3066.25)(0.40)/(1750)

w[tex]_{}2[/tex]= 0.70 rad/s [which is to be expected, however apparently incorrect]

At this point, I realized that for some reason the conservation of angular momentum must not apply in this scenario, and so I did

[tex]\theta[/tex] = 0degrees (the angle between the radius and the applied force, assuming they jump of radially)

T=rfsin[tex]\theta[/tex]=0

[tex]\alpha[/tex]=T/I

[tex]\alpha[/tex]=0

Therefore [tex]\Delta[/tex]w=0 [the correct answer]

Essentially, my question is why the law of conservation of angular momentum applies when people step onto the merry-go-round, but not when they step off? I know I am making some sort of obvious logical blunder, but I cannot see it :(.