A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 3.4 kg weight is hanging on the string. The system of the weight and disk is released from rest. 1)When the 3.4 kg weight is moving with a speed of 1.1 m/s, what is the kinetic energy of the entire system? K. E. =

5.45 J

Explanation:

When the 3.4 kg weight is moving with a speed of 1.1 m/s, what is the kinetic energy of the entire system?

RKE =

where;

RKE =

=

= 3.75 J

LKE =

=

= 1. 7 J

K.E = RKE + LKE

K. E = ( 3.75 + 1.7 ) J

K . E = 5.45 J

The answer is 6.6Joules

Explanation:

The kinetic energy of the entire system = the rotational kinetic energy of the disk + the kinetic energy of the body.

That is, K.E = RKE + BKE

The formula of a rotational kinetic energy is given as:

   RKE = Rotational Kinetic energy,

    I =     Inertia of the disk

     w = angular velocity

for inertia the formula is given as , I =----------------2

where:

m = mass and r is the radius

Substituting the values into equation 2, we have

I = 1/2

 =1/2* 15 * 0.25^2

= 1/2 *15 * 0.0625

= 0.46875

Also for w which is angular velocity, we have

w = v/r-----------------------------------------------3

Where v is the velocity and r is the radius.

Substituting the values into equation 3, we have

w = v/r

  = 1.1/0.25

  = 4.4rads

Now putting the calculated values into the main equation 1, we have

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