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

Question

3 years ago

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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. =

Physics

2 answers:

natulia [17] · Answer 1 · 2022-08-10T03:14:52+03:00

Answer:

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:

<h2>RKE = $1/2Iw^{2}$ -----------------------1</h2><h2>Where: </h2><h2 />

RKE = Rotational Kinetic energy,

I = Inertia of the disk

w = angular velocity

for inertia the formula is given as , I = $mr^{2}$ ----------------2

where:

m = mass and r is the radius

Substituting the values into equation 2, we have

I = 1/2 $mr^{2}$

=1/2* 15 * 0.25^2

= 1/2 *15 * 0.0625

= 0.46875 $kgm^{2}$

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

<h2>RKE = $1/2Iw^{2}$ </h2><h2> = 1/2 * 0.9375 * 4.4^2</h2><h2> = 1/2 * 0.46875 * 19.36</h2><h2> = 1/2 * 9.075</h2><h2> = 4.5375J</h2><h3>Calculating the kinetic energy of the body BKE, we have:</h3><h3>BKE = 1/2mv^2</h3><h3> 1/2 * 3.4 * 1.1^2</h3><h3> = 1/2 * 3.4 * 1.21</h3><h3> = 1/2 * 4.114</h3><h3> = 2.057 J</h3><h3>Therefore the kinetic energy of the entire system = RKE + BKE = 4.5375 + 2.057 = 6.59J = 6.6J approximately</h3>