Question

The string constrains the rotational and translational motion of the falling cylinder, given that it doesn't slip. what is the relationship between the magnitude of the angular velocity ω and that of the velocity v of the center of mass of the cylinder? express ω in terms of v and r.

Answer 1

The angular velocity of the rolling cylinder is expressed as $\boxed{\omega=\frac{{2gt}}{{3r}}}$ .

Further Explanation:

Since the cylinder rolls along with the string which is wrap around it. There is a tension developed in the string due to weight of the cylinder and its acceleration.

The tension developed in the string is given as:

$\begin{aligned}T&=mg- ma\\&= m\left({g - a}\right)\\\end{aligned}$

This tension developed in the string produces a torque about the center which makes the cylinder to roll. The torque developed about center is expressed as:

$\begin{aligned}\tau&=T\timesr\\&= m\left({g - a} \right)r\\\end{aligned}$

The torque developed in a rotating body is also expressed as:

 $\tau = I\alpha$

Here, $I$ is the moment of inertia of the cylinder and $\alpha$ is the angular acceleration of the cylinder.

The moment of inertia of a cylinder is $\dfrac{1}{2}m{r^2}$ and the angular acceleration for a acceleration body can also be expressed as $\alpha =\dfrac{a}{r}$.

Thus comparing the expressions of torque:

$\begin{alinged}\dfrac{1}{2}m{r^2}\times\frac{a}{r}&= m\left({g - a}\right)r\hfill\\a&=\dfrac{{2g}}{3}\hfill\\\end{aligned}$

The linear velocity of the cylinder after time $t$ will be:

 $\begin{aligned}v&= a\times t\\&=\frac{{2gt}}{3}\\\end{aligned}$

The angular velocity of a rotating body is expressed as:

  $\begin{aligned}\omega&= \frac{v}{r}\\&=\frac{{2gt}}{{3r}}\\\end{aligned}$

Thus, the angular velocity of the rolling cylinder is expressed as $\boxed{\omega=\frac{{2gt}}{{3r}}}$.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Rotational Motion

Keywords:  String constrains, rotational motion, translational motion, angular velocity, falling cylinder, torque, center of mass of cylinder, linear speed of center of mass.

Answer 2

Answer: First we need to find the acceleration. torque on cylinder τ = T * r where T is the string tension; T = m(g - a) where a is the acceleration of the cylinder. Then τ = m(g - a)r But also τ = Iα. For a solid cylinder, I = ½mr², and if the string doesn't slip, then α = a / r, so τ = ½mr² * a/r = ½mra. Since τ = τ, we have m(g - a)r = ½mra → m, r cancel, leaving g - a = ½a g = 3a/2 a = 2g/3 where g, of course, is gravitational acceleration. We know that v(t) = a*t, so for our cylinder v(t) = 2gt / 3 ◄ linear velocity and ω = v(t) / r = 2gt / 3r ◄ angular velocity