Answer: a) 24.2 rad/s² b) 6.95 rad/s
Explanation:
So the cylinder's moment of inertia is [tex]I = \frac{MR^2}{2}[/tex], where M is the mass of the cylinder and R is its radius.
The problem you have here is that the axle is modifying it, so we use the parallel axis theorem:
Therefore
[tex]I = \frac{MR^2}{2} + MD^2[/tex] where D is the distance from the center of mass to the axis which is a radius R
[tex]\=> I = \frac{MR^2}{2} + M(R)^2 = \frac{3MR^2}{2}[/tex]
Next the Momentum with respect to the center of mass of the cylinder is
[tex]M_{c} = I\alpha = F*R = m*g*R[/tex] because there are no other forces rather than the cylinder's weight, now [tex]I = \frac{3*5*0.27^2}{2} = 0.54675[/tex] kg*m^2
a ) replace and find [tex]\alpha = (m*g*R)/(I ) [/tex]
[tex]\alpha = (5.0*9.8*0.27)/(0.54675 ) = 24.2 rad/s^2[/tex]
b) we start by assuming that the energy is conserved because the cylinder is rolling without sliding:
so the potential energy is given by [tex]E = M*g*R = 5*9.8*0.27 = 13.23J[/tex]
now you have that the same amount is going to be equal to the kinetic rotational energy of the cylinder so [tex]13.23J = \frac{I*\omega^2}{2} => \omega = \sqrt(\frac{2*13.23J}{0.54675} ) = 6.95 rad/s[/tex]
(a) The angular acceleration of the cylinder is 24.1 rad/s².
(b) e magnitude of the cylinder's angular velocity is 6.94 rad/s.
The moment of inertia of the cylinder can be determined using parallel axis theorem as shown below;
I = ³/₂MR²
I = ³/₂ x 5 x (0.27)²
I = 0.55 kgm²
The angular acceleration of the cylinder is calculated as follows;
Iα = Fr
α = Fr/I
α = (mgr)/I
α = (5 x 9.8 x 0.27)/0.55
α = 24.1 rad/s²
The potential energy of the cylinder which is converted into rotational kinetic energy is calculated as follows;
P.E = mgh
P.E = mgR
P.E = 5 x 9.8 x (0.27)
P.E = 13.23 J
[tex]K.E =P.E = \frac{1}{2} I \omega ^2\\\\\omega^2 = \frac{2K.E}{I} \\\\\omega = \sqrt{\frac{2K.E}{I} } \\\\\omega = \sqrt{\frac{2(13.23)}{0.55} }\\\\\omega = 6.94 \ rad/s[/tex]
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