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

`\alpha=58.01\ rad.s^(-1)`

Step-by-step 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=(1)/(2) m.r^2`

`I=0.5* 2.39* 0.123^2`

`I=0.0181\ kg.m^2`

We know the torque is given as:

`\tau=F_T* r`

`\tau=8.535* 0.123`

`\tau=1.05\ N.m`

Now, also

`\tau=I.\alpha`

`1.05=0.0181* \alpha`

`\alpha=58.01\ rad.s^(-1)`

Answer 2

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

Step-by-step 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* r=(1)/(2)mr^2*\alpha`

Where, F = force

r = radius

m = mass

Put the value into the formula

`8.535*0.123=(1)/(2)*2.39*(0.123)^2*\alpha`

`\alpha=(2*8.535*0.123)/(2.39*(0.123)^2)`

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

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