Rotational Motions Questions

Questions

A $12\, {\rm cm}$ diameter, $2\,{\rm kg}$ uniform circular disk, which is initially at rest, experiences the net torque shown in the figure below. What is the disk's angular velocity at $t=12\, \mathrm{s}$? The disk rotates about an axis perpendicular to the plane of the disk and through its center. Note: $I_{disk}=\frac{1}{2}MR^2$

A uniform rectangular beam of length $L=5\, \mathrm{m}$and mass $M=40\, \mathrm{kg}$ is supported but not attached to the two posts which are length $D=3\, \mathrm{m}$ apart. A child of mass $W=20$ kg starts walking along the beam.
(a) Assuming infinitely rigid posts, how close can the child get to the right end of the beam without it falling over? [Hint: The upward force exerted by the left on the beam cannot be negative --this is the limiting condition on how far the child can be to the right. Set up the static equilibrium condition with the pivot about the left end of the beam.]
(b) Suppose the left end of the beam is attached to the left post although it can still freely pivot about it. The right post is not attached as in part a). If the left post is infinitely rigid while the Young's modulus for the right post which is of length $s=10\, {\rm m}$ is $Y={10}^{10}\, \mathrm{N/}{\mathrm{m}}^{\mathrm{2}}$ and its cross sectional area is $A=\frac{1}{2}\,{\mathrm{m}}^{\mathrm{2}}$, how much does its length differ between the situation when the child is not present and when the child is at the rightmost edge of the beam?

A grindstone spinning at the rate of $8.3\ \mathrm{rev/s}$ has what approximate angular speed?

The figure below represents a $2.0\, {\rm m}$ long bar pinned at point A. compute the net torque on the bar with respect to point A due to the forces $F_1=50\, \mathrm{N}$ applied 1.0 m from point A, and $F_2=100\,\mathrm{N}$ as shown in the figure below.

A physical pendulum has a shape of a disk of radius $r$ and mass $m$. The pendulum swings about an axis perpendicular to the plane of the disk and at a distance $L$ from the center of the disk.
(a) What is the moment of inertia of the disk about the axis where the pendulum swings?
(b) What is the oscillation frequency of this pendulum?
(c) For what value of $L$ is this frequency at a maximum?
($I_{disk}$ about its center of mass is $\frac{1}{2}mr^2$- assume that the oscillation amplitude is small and small angle approximation is ${\sin \theta \ }\sim \theta$ and ${\cos \theta \ }\sim 1-\frac{{\theta }^2}{2}$ )

(a) Note: parallel axis theorem
The moment of inertia is minimal when the rotation axis passes through the center-of-mass(CM) and increases as the rotation axis is moved further from the (CM). i.e. $I=I_{cm}+mh^2$, where $h$ is the distance from the CM to the parallel axis of rotation.
So, $I=\frac{1}{2}mr^2+mL^2$
(b) The radial part of $mg$ will be balanced by the tension on the cord (so there is no radial acceleration). The tangential part of it provides the torque on the pendulum.
$\vec{\tau }=I\alpha =I\frac{d^2\theta }{dt^2}\ \Rightarrow \ -mgL\,{\sin \theta \ }=I\frac{d^2\theta }{dt^2}$
$\therefore \frac{d^2\theta }{dt^2}+\frac{mgL}{I}{\sin \theta \ }=0$
$\mathrm{small\ angle\ Approx.}\ \ :\ \frac{d^2\theta }{dt^2}+\frac{mgL}{I}\theta \approx 0$
Note: In above the minus sign is there because the torque opposes the angular displacement from equilibrium.

by considering $\frac {mgL}{I}$ as the squared angular momentum $\omega^2$, the above equation have similar form of an SHM equation, therefore
${\omega }^2=\frac{mgL}{I}=\frac{mgL}{mL^2+\frac{1}{2}mr^2} \quad , \quad f=\frac{\omega }{2\pi }$

(c) To find the extremes of a function, we must take the derivative of it.If $\omega$ is an extreme, so ${\omega }^2$ is also, therefore we can examine $d{\omega }^2/dL$ rather $d\omega /dL$
$\frac{d{\omega }^2}{dL}=1-\frac{r^2}{L^2}\frac{1}{1+r^22L^2}=0\ \Longrightarrow \ \ L=R/\sqrt{2}$
It is less than $R$!

Two astronauts, each having a mass $M$, are connected by a rope of length $d$ having negligible mass. They are isolated in space, orbiting their central mass at speeds $v$. Treating the astronauts as particles, calculate
(a) The angular momentum of the system
(b) The rotational energy of the system
By pulling on the rope, one of astronauts shortens the distance between them to $d/2$.
(c) What is the new angular momentum of the system?
(d) What are the astronauts' new speeds?
(e) What is the new rotational energy of the system?
(f) How much work does the astronaut do in shortening the rope?

The upper end of the string wrapped around the cylinder in the figure below is held by a hand that is accelerated upward so that the center of mass of the cylinder does not move as the cylinder spins up. Find:
(a) The tension in the string
(b) The angular acceleration of the cylinder
(c) The acceleration of the hand.

A force $\vec{F}=3\hat{i}-2\hat{j}\ \mathrm{(N)}$ acts at a location $\vec{r}=1\ \hat{i}+2\hat{j}\ \mathrm{(m)}$ on an object. What is the torque that this force applies about an axis through the origin perpendicular to the $xy$ plane?

A man of mass $M=75\,{\rm kg}$ lowers himself down from the top of a building by using a rope wound on a drum (a hollow cylinder of radius $r=0.50\,{\rm m}$ and mass $2M=150\, {\rm kg}$), as shown in the picture. The man and the drum start at rest.
(a) Find the angular acceleration of the drum.
(b) What is the velocity of the man when he has dropped $\mathrm{20\ m}$?

A hoop is released from rest at the top of a plane inclined at $20{}^\circ$ above horizontal. How long does it take the hoop to roll $12.0\ \mathrm{m}$ down the plane? ($I_{hoop}=mr^2$)

In this problem we have a rotational motion. So use the conservation of mechanical energy between initial and final points to find the velocity of the center of mass of the hoop at the bottom of the incline plane.
$E_{top}=E_{bottom}$
$mgh=\underbrace{\frac{1}{2}mv^2_c}_{ \begin{array}{c} trans.\ \ kinetic \\ energy \end{array} }+\underbrace{\frac{1}{2}I{\omega }^2}_{ \begin{array}{c} rotational\ kinetic \\ energy \end{array} }$
We know that in the rotational motions, the angular and tangential velocities are related together via $v=r\omega$.
By substituting the given values, we obtain
$mgh=\frac{1}{2}mv^2_c+\frac{1}{2}\left(mr^2\right){\left(\frac{v_c}{r}\right)}^2\Rightarrow v_c=\sqrt{gh}=\sqrt{g(L\,{\sin 20{}^\circ \ })}$
From the geometry, we have substituted $L\,{\sin 20{}^\circ \ }$ for height of the incline plane $h$.
Now using the kinematic equation $v^2-v^2_0=2ax$, find the acceleration of the CM and then use the equation $y=\frac{1}{2}at^2+v_0t$ to determine the required time.
$v^2-\underbrace{v^2_0}_{0}=2a\underbrace{x}_{L}\Rightarrow \ a=\frac{gL}{2L}{\sin 20{}^\circ \ }$
If we suppose the axis parallel to the incline plane as the $y$ axis then $y=L$
$y=\frac{1}{2}at^2+\underbrace{v_0}_{0}t\to L=\frac{1}{2}at^2$
\begin{align*}
\Rightarrow t & =\sqrt{\frac{2L}{a}}=\sqrt{\frac{2L}{\left(\frac{gL}{2L}\,{\sin 20{}^\circ \ }\right)}}=\sqrt{\frac{4L}{g{\sin 20{}^\circ \ }}}\\
& =\sqrt{\frac{4\left(12\right)}{9.8\,{\sin 20{}^\circ \ }}}=3.78\ \mathrm{s}
\end{align*}

A ball begins rolling up a hill as shown, with no slipping or loss of mechanical energy. Its initial velocity in the horizontal flat region is $30\ \mathrm{m/s}$. The moment of inertia of a solid sphere with mass $M$ and radius $R$ is given by $I=\frac{2}{5}MR^2$.
(a) Calculate its velocity $v_f$ just as it rolls horizontally off the edge of the cliff, which is $20\ \mathrm{m}$ above the initial position.
(b) Calculate the time $t$ that it takes to hit the ground after falling off the cliff.
(c) Calculate the distance from the base of the cliff to the point where the ball hits the ground.
(d) Calculate the (center of mass) velocity that the ball has when it hits the ground.

A block of mass $m$ is being whirled around on a frictionless surface (as shown), in a circular path of radius $r$.
(a) Its initial velocity is $v$. If it is pulled in so that the new radius is $r/2$, what is its new velocity?
(b) The initial tension in the string is $T$. What is the final tension after the block is pulled into the new radius of $r/2$?

A solid disk has a radius of $R=0.1\ \mathrm{m}$ and mass of $M=2.0\ \mathrm{kg}$. it begins rolling up a slope without slipping.  The slope is $\theta =20{}^\circ$ above the horizontal, and the disk has an initial speed of $v_0=3.0\ \mathrm{m/s}$. How long it will take for the disk to come to a stop?

Consider the masses configured on a frictionless inclined plane as shown in the figure. The pulley is frictionless, has a moment of inertia $I$ and a radius $R$. Determine the correct expression for the acceleration of $m_1$ in terms of the other quantities. Take the positive direction for the acceleration of $m_1$ to be down.

A uniform disk of mass $M=10\ \mathrm{kg}$ and radius $R=0.500\ \mathrm{m}$ is rolling without slipping on a horizontal surface. The rolling disk has a linear (or translational) velocity of $v_{cm}=5\ \mathrm{m/s}$ to the right, as shown in the figure. The moment of inertial of the disk rotating about its center of mass is $I=\frac{1}{2}MR^2$.
(a) What is the angular velocity of the rolling disk about its center of mass?
(b) What is the kinetic energy of the rolling disk?
(c) If this disk rolled up a hill, what is the maximum height (above its starting point) that it would reach?
(d) If instead of a rolling disk, you were given a square sliding mass ($m=10\ \mathrm{kg}$) with the same initial linear velocity $v_{cm}=5\ \mathrm{m/s}$, what maximum height could the sliding mass reach? (assuming all surfaces were frictionless)

(a) The angular velocity $\omega$ of a rolling object is related to its tangential velocity by $v=R\omega$. Thus
$v_{cm}=R\omega \to \omega =\frac{v_{cm}}{R}=\frac{5}{0.5}=10\ \mathrm{rad/s}$
(b) A rolling object has two kinetic energy, translational and rotational. Therefore,
$K_{tot}=K_{tran}+K_{rot}=\frac{1}{2}Mv^2_{cm}+\frac{1}{2}I_{cm}{\omega }^2=\frac{1}{2}Mv^2_{cm}+\frac{1}{2}\left(\frac{1}{2}MR^2\right){\omega }^2$
$\Rightarrow K_{tot}=\frac{1}{2}\left(10\right){\left(5\right)}^2+\frac{1}{4}\left(10\right){\left(0.5\right)}^2{\left(10\right)}^2=188\ \mathrm{J}$
(c) Use the conservation of the mechanical energy as $E_i=E_f$.
$U_i+K_i=U_f+K_f\to \ Mgh_i+K_{i,tot}=MgH+K_{f,tot}$
Let the lowest point of the hill be the base of the gravitational potential so $U_i=0$. At the final point the rolling object comes to a stop, $K_{f,tot}=0$. Therefore
$0+188=\left(10\right)\left(9.8\right)H+0\to H=\frac{188}{98}=1.92\mathrm{\ m}$
(d) Use again the conservation of mechanical energy, but in this case, there is no rotational kinetic energy so
$E_i=E_f\to mgh_i+\frac{1}{2}mv^2_{cm}=mgH+K_f$
$\Rightarrow 0+\frac{1}{2}mv^2_{cm}=mgH+0\Rightarrow H=\frac{v^2_{cm}}{2g}=\frac{5^2}{2\left(9.8\right)}=1.28\ \mathrm{m}$

A uniform density metal plate in the shape of an equilateral triangle with each side of length$L=2.00\ \mathrm{m}$, is pivoted about an axis perpendicular to the plate and located at its center. Two small rocket engines are attached to the tips of the triangle that act to create two forces on the triangle $F_1=100\ \mathrm{N}$ and $F_2=100\ \mathrm{N}$ as shown in the figure.
(a) Calculate the torque about the pivot axis from the force $F_1$.
(b) Calculate the torque about the pivot axis from force $F_2$.
(c) Calculate the total(or net) torque about the pivot axis.
(d) You now do an experiment and find that, starting from rest, the triangle rotates about its center through one complete revolution in $4.00$ seconds. Using this information, calculate the moment of inertia about its center.

A cylindrical log of mass $M$ rolls without slipping along the ground. Its center moves along at speed $V$. What is the kinetic energy of the log in terms of $M$ and $V$?

Category : Rotational Motions

MOST USEFUL FORMULA IN ROTATIONAL MOTION:

Relationship between linear and angular speeds:
$v=r\omega$
Angular speed $\omega$ must be measured in radians per second ($\mathrm {rad/s}$).

Tangential acceleration of a point on a rotating object:
$a_{tra}=\frac {dv}{dt}=r\frac {d\omega}{dt}=r\alpha$

Definition of moment of inertia:
$I=m_1r_1^2+m_2r_2^2+\dotsc=\Sigma m_ir_i^2$

Kinetic energy of a rotating body:
$K=\frac 1 2 I\omega^2$

Parallel-Axis theorem:
$I_P=I_{cm}+Md^2$
where $d$ is the distance between cenetr of mass and the axis of rotation.

Definition of torque:
$\vec \tau=\vec r \times \vec F$

Rotational analog of Newton's second law:
$\Sigma \tau=I\alpha$

Angular momentum of a particle w.r.t a point:
$\vec L=\vec r\times \vec p=\vec r\times m\vec v$

Angular momentum of a rotating body around an axis:
$\vec L=I\vec \omega$

Number Of Questions : 17