Respuesta :
Answer:
First we need to find the acceleration.
torque on cylinder Ď„ = T * r where T is the string tension;
T = m(g - a) where a is the acceleration of the cylinder. Then
Ď„ = 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 Ď„ = Ď„, 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 â—„ linear velocity
and ω = v(t) / r = 2gt / 3r ◄ angular velocity
The angular velocity of the rolling cylinder is expressed as [tex]\boxed{\omega=\frac{{2gt}}{{3r}}}[/tex] .
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:
[tex]\begin{aligned}T&=mg- ma\\&= m\left({g - a}\right)\\\end{aligned}[/tex]
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:
[tex]\begin{aligned}\tau&=T\timesr\\&= m\left({g - a} \right)r\\\end{aligned}[/tex]
The torque developed in a rotating body is also expressed as:
[tex]\tau = I\alpha[/tex]
Here, [tex]I[/tex] is the moment of inertia of the cylinder and [tex]\alpha[/tex] is the angular acceleration of the cylinder.
The moment of inertia of a cylinder is [tex]\dfrac{1}{2}m{r^2}[/tex] and the angular acceleration for a acceleration body can also be expressed as [tex]\alpha =\dfrac{a}{r}[/tex].
Thus comparing the expressions of torque:
[tex]\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}[/tex]
The linear velocity of the cylinder after time [tex]t[/tex] will be:
[tex]\begin{aligned}v&= a\times t\\&=\frac{{2gt}}{3}\\\end{aligned}[/tex]
The angular velocity of a rotating body is expressed as:
[tex]\begin{aligned}\omega&= \frac{v}{r}\\&=\frac{{2gt}}{{3r}}\\\end{aligned}[/tex]
Thus, the angular velocity of the rolling cylinder is expressed as [tex]\boxed{\omega=\frac{{2gt}}{{3r}}}[/tex].
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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.
