Question

A student sits on a rotating stool holding two 4.0 kg objects.
When his arms are extended horizontally, the objects are 1.0 m from
the axis of rotation, and he rotates with an angular speed of 0.75
rad/s. The moment of inertia of the student plus stool is 3.0
kg·m^{2} and is assumed to be constant. The student then
pulls the objects horizontally to 0.20 m from the rotation
axis.

(a) Find the new angular speed of the student.

rad/s

(b) Find the kinetic energy of the student before and after the
objects are pulled in.

before | J |

after | J |

Answer #1

m = 2 kg ; r1 = 1 m ; I = 3 kg-m^2 ; w = 0.75 rad/s ;

r2 = 0.2 m

a)The total moment of inertia of the student and object system is:

I = I(stool+student) + 2 x I(object) = 3 + 2 m r^2

I1 = 3 kg-m^2 + 2 x 2 x 1^2 = 7 kg-m^2

later I becomes:

I2 = 3 + 2 x 2 x 0.2^2 = 3.16 kg-m^2

we know from conservation of angular momentum that

Li = Lf

I1w1 = I2w2

w2 = I1w1/I2 = 7 x 0.75/3.16 = 1.66 rad/s

**Hence, w2 = 1.66 rad/s**

b)Before KE is:

KE(before) = 1/2 I w^2

KE(before) = 0.5 x 7 x 0.75 = 1.97 J

KE(after) = 0.5 x 3.16 x 1.66 x 1.66 = 4.35 J

**Hence, KE(before) = 1.97 J and KE(after) = 4.35
J**

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