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November 24, 2014

November 24, 2014

Posted by **BB** on Monday, February 27, 2012 at 12:30pm.

A light rope is wrapped several times around a large wheel with a radius of 0.435 m. The wheel rotates in frictionless bearings about a stationary horizontal axis. The free end of the rope is tied to a suitcase with a mass of 18.0 kg. The suitcase is released from rest at a height of 4.00 m above the ground. The suitcase has a speed of 3.60 m/s when it reaches the ground.

(a) the angular velocity of the wheel when the suitcase reaches the ground

my correct anwser is 8.28 rad/s

(b) the moment of inertia of the wheel.

I=? kg*m^2

im having trouble with just part b

- Physics help -
**Elena**, Monday, February 27, 2012 at 6:02pma. ω = v/R = 3.6/0.435 = 8.28 rad/s2

b. From kinematics:

s = v0 - a•t^2/2

v = v0 + a•t.

Since v0 = 0, a = v^2/2•s =

= ( 3.6)^2/2•4 = 1.62 m/s2.

Second law:

m•a = m•g – T,

T = m•g - m•a= 18(9.8-1.62) =147.24 N.

The second law for rotation:

I•ε = M

I• (a/R)=T•R,

I=T•R^2/a=147.24•(0.435)^2/1.62=

=17.2 kg•m2.

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