Question

M, a solid cylinder (M=2.39 kg, R=0.123 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.870 kg mass, i.e., F = 8.535 N. Calculate the angular acceleration of the cylinder.

Answer 1

Answer:

$\alpha=58.01\ rad.s^{-1}$

Explanation:

Given:

• mass of solid cylinder, $m=2.39\ kg$

• radius of solid cylinder, $r=0.123\ m$

• tangential force on the solid cylinder, $F_T=8.535\ N$

Moment of inertial of a solid cylinder:

$I=\frac{1}{2} m.r^2$

$I=0.5\times 2.39\times 0.123^2$

$I=0.0181\ kg.m^2$

We know the torque is given as:

$\tau=F_T\times r$

$\tau=8.535\times 0.123$

$\tau=1.05\ N.m$

Now, also

$\tau=I.\alpha$

$1.05=0.0181\times \alpha$

$\alpha=58.01\ rad.s^{-1}$

Answer 2

Answer:

The angular acceleration of the cylinder is 58.06 rad/s².

Explanation:

Given that,

Mass $M=2.39\ kg$

Radius $R=0.123\ m$

Force $F=8.535\ N$

Weight $W=0.870\ kg$

We need to calculate the angular acceleration of the cylinder

Using formula of torque

$\tau=I\alpha$

$F\times r=\dfrac{1}{2}mr^2\times\alpha$

Where, F = force

r = radius

m = mass

Put the value into the formula

$8.535\times0.123=\dfrac{1}{2}\times2.39\times(0.123)^2\times\alpha$

$\alpha=\dfrac{2\times8.535\times0.123}{2.39\times(0.123)^2}$

$\alpha=58.06\ rad/s^2$

Hence, The angular acceleration of the cylinder is 58.06 rad/s².