Respuesta :
Answer:
The angular acceleration of the cylinder is 58.06 rad/s².
Explanation:
Given that,
Mass [tex]M=2.39\ kg[/tex]
Radius [tex]R=0.123\ m[/tex]
Force [tex]F=8.535\ N[/tex]
Weight [tex]W=0.870\ kg[/tex]
We need to calculate the angular acceleration of the cylinder
Using formula of torque
[tex]\tau=I\alpha[/tex]
[tex]F\times r=\dfrac{1}{2}mr^2\times\alpha[/tex]
Where, F = force
r = radius
m = mass
Put the value into the formula
[tex]8.535\times0.123=\dfrac{1}{2}\times2.39\times(0.123)^2\times\alpha[/tex]
[tex]\alpha=\dfrac{2\times8.535\times0.123}{2.39\times(0.123)^2}[/tex]
[tex]\alpha=58.06\ rad/s^2[/tex]
Hence, The angular acceleration of the cylinder is 58.06 rad/s².
Answer:
[tex]\alpha=58.01\ rad.s^{-1}[/tex]
Explanation:
Given:
- mass of solid cylinder, [tex]m=2.39\ kg[/tex]
- radius of solid cylinder, [tex]r=0.123\ m[/tex]
- tangential force on the solid cylinder, [tex]F_T=8.535\ N[/tex]
Moment of inertial of a solid cylinder:
[tex]I=\frac{1}{2} m.r^2[/tex]
[tex]I=0.5\times 2.39\times 0.123^2[/tex]
[tex]I=0.0181\ kg.m^2[/tex]
We know the torque is given as:
[tex]\tau=F_T\times r[/tex]
[tex]\tau=8.535\times 0.123[/tex]
[tex]\tau=1.05\ N.m[/tex]
Now, also
[tex]\tau=I.\alpha[/tex]
[tex]1.05=0.0181\times \alpha[/tex]
[tex]\alpha=58.01\ rad.s^{-1}[/tex]
