Answer :
Answer:
(a). The magnitude of the torque exerted by the rope is 2.6 N-m
(b). The angular acceleration of the cylinder is 109.8 rad/s².
(c). The angular velocity of the cylinder is 131.76 rad/sec.
Explanation:
Given that,
Mass of cylinder = 2.8 kg
Radius = 13 cm
Force = 20 N
(a). We need to calculate the magnitude of the torque exerted by the rope
Using formula of torque
[tex]\tau=F\times r[/tex]
Put the value into the formula
[tex]\tau=20\times13\times10^{-2}[/tex]
[tex]\tau=2.6\ N-m[/tex]
The magnitude of the torque exerted by the rope is 2.6 N-m
(b). We need to calculate the angular acceleration of the cylinder
Using formula of acceleration
[tex]\alpha=\dfrac{\tau}{I}[/tex]
[tex]\alpha=\dfrac{2\tau}{mr^2}[/tex]
[tex]\alpha=\dfrac{2\times2.6}{2.8\times(13\times10^{-2})^2}[/tex]
[tex]\alpha=109.8\ rad/s^2[/tex]
The angular acceleration of the cylinder is 109.8 rad/s².
(c). We need to calculate the angular velocity of the cylinder
Using equation of motion
[tex]\omega=\omega_{0}+\alpha t[/tex]
Put the value into the formula
[tex]\omega=109.8\times1.20[/tex]
[tex]\omega=131.76\ rad/sec[/tex]
The angular velocity of the cylinder is 131.76 rad/sec.
Hence, (a). The magnitude of the torque exerted by the rope is 2.6 N-m
(b). The angular acceleration of the cylinder is 109.8 rad/s².
(c). The angular velocity of the cylinder is 131.76 rad/sec.