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# A solid cylindrical disk whose mass is 1.4 kg and radius is 8.5 cm, rolls across a horizontal table at a speed of 15 cm/...

A solid cylindrical disk whose mass is 1.4 kg and radius is 8.5 cm, rolls across a horizontal table at a speed of 15 cm/s. What is the instantaneous velocity of the top of the rolling cylinder? What is the angular speed of the rolling disk? What is the kinetic energy of the rolling disk?

Given:

Mass of the disk is $1.4kg$.

Radius of the disk is $r=8.5cm=8.5×{10}^{-2}m$.

Rolls across a horizontal table at speed ${\stackrel{⇀}{v}}_{com}=15cm}{s=15×{10}^{-2}m}{s}}$.

Calculation:

The instantaneous velocity at the top of the rolling cylindrical disk is given as:

$\stackrel{⇀}{v}=2{\stackrel{⇀}{v}}_{com}$

Here, ${\stackrel{⇀}{v}}_{com}$ is velocity of center of mass given as $15×{10}^{-2}m}{s}$. So,

$\stackrel{⇀}{v}=2\left(15×{10}^{-2}m}{s}\right)\phantom{\rule{0ex}{0ex}}\stackrel{⇀}{v}=0.30m}{s}=30cm}{s}$

Thus, the instantaneous velocity of the top of the rolling cylinder is $30cm}{s}$.

The angular speed of the rolling disk is given as:

$\omega =\frac{{v}_{com}}{r}$

Substitute  for $8.5cm$ for $r$ and $15cm}{s}$ for ${\stackrel{⇀}{v}}_{com}$.

$\omega =\frac{15cm}{s}}{8.5cm}\phantom{\rule{0ex}{0ex}}\omega =1.7647rad}{s}$

Thus, the angular speed of the rolling disk is $1.7647rad}{s}$.

The kinetic energy for the rolling disk is given as:

Here, ${I}_{com}$ is moment of inertia about center of mass given as:

Now, kinetic energy:

Thus, the kinetic energy of the rolling disk is $23.62×{10}^{-3}J$

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