Answer :
Answer:
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 = [tex]\frac{1}{2}I \omega ^2[/tex]
where;
[tex]I = \frac{1}{2} mr^2[/tex]
[tex]I = \frac{1}{2}*15*0.25^2[/tex]
[tex]I = 0.46875 kg.m^2[/tex]
[tex]\omega = \frac{1.0}{0.25}[/tex]
[tex]\omega = 4 rad/s[/tex]
RKE = [tex]\frac{1}{2}I \omega ^2[/tex]
= [tex]\frac{1}{2} *0.46875*4^2[/tex]
= 3.75 J
LKE = [tex]\frac{1}{2} mv^2[/tex]
= [tex]\frac{1}{2} *3.4*1.0^2[/tex]
= 1. 7 J
K.E = RKE + LKE
K. E = ( 3.75 + 1.7 ) J
K . E = 5.45 J
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:
RKE = [tex]1/2Iw^{2}[/tex]-----------------------1
Where:
RKE = Rotational Kinetic energy,
I = Inertia of the disk
w = angular velocity
for inertia the formula is given as , I =[tex]mr^{2}[/tex]----------------2
where:
m = mass and r is the radius
Substituting the values into equation 2, we have
I = 1/2[tex]mr^{2}[/tex]
=1/2* 15 * 0.25^2
= 1/2 *15 * 0.0625
= 0.46875[tex]kgm^{2}[/tex]
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
RKE = [tex]1/2Iw^{2}[/tex]
= 1/2 * 0.9375 * 4.4^2
= 1/2 * 0.46875 * 19.36
= 1/2 * 9.075
= 4.5375J
Calculating the kinetic energy of the body BKE, we have:
BKE = 1/2mv^2
1/2 * 3.4 * 1.1^2
= 1/2 * 3.4 * 1.21
= 1/2 * 4.114
= 2.057 J
Therefore the kinetic energy of the entire system = RKE + BKE = 4.5375 + 2.057 = 6.59J = 6.6J approximately