Question

A 5.0 kg, 60-cm-diameter disk rotates on an axle passing throughone edge.The axle is parallel to the floor. The cylinder isheld with the center of mass at thesame height as the axle, thenreleased.
What is the cylinder's initial angularacceleration? in rad/s^2What is the cylinder's angular velocity when itis directly below the axle? in rad/s
$knight_Figure_13_72.jpg$

Answer 1

a. Moment of Inertia = Icm = m r^2/2

Parallel Axis Theorem I = Icm + md^2

r = 30 cm

d = 30 cm

I = mr^2/2 + mr^2 = 3/2 mr^2

F = mg

Torque = τ = Fx d = mgr = I α

α = mgr/I = 2g/3r

Answer 2

(a) The only force acting on the disk is its weight \(\therefore\) Force acting on the disk \(F=m g\) \(m\) Is mass of the disk \(=5 \mathrm{~kg}\) \(g\) Is acceleration due to gravity \(=9.8 \mathrm{~m} / \mathrm{s}^{2}\) Then \(F=(5 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)=49 \mathrm{~N}\)

Then torque acting on the wheel about the axis is \(\tau=F \times l\) is distance from the axis to its centre of mass this is nothing that the radius of the disk \(l=\frac{D}{2}\) \(D\) is diameter of the disk \(=60 \mathrm{~cm}=0.6 \mathrm{~m}\) \(\therefore l=\frac{0.6 \mathrm{~m}}{2}=0.3 \mathrm{~m}\)

Then \(\tau=(49 \mathrm{~N})(0.3 \mathrm{~m})=14.7 \mathrm{Nm}\)

Also torque \(\tau=I \alpha\) I is moment of inertia of the disk about the axle \(=I_{c}+M R^{2}\) \(I_{c}\) is the moment of inertia of the flywheel about an axis passing through The centre of mass and perpendicular to it \(=\frac{1}{2} M R^{2}\)

$$ \begin{aligned} \therefore I &=\frac{1}{2} M R^{2}+M R^{2} \\ &=\frac{3}{2} M R^{2}=\left(\frac{3}{2}\right)(5 \mathrm{~kg})(0.3 \mathrm{~m})^{2} \\ &=0.675 \mathrm{~kg} \mathrm{~m}^{2} \end{aligned} $$

Then angular acceleration \(\alpha=\frac{\tau}{I}\)

$$ =\frac{(14.7 \mathrm{Nm})}{\left(0.675 \mathrm{~kg} \mathrm{~m}^{2}\right)}=21.78 \mathrm{rad} / \mathrm{s}^{2} $$

(b) Now from the initial position to a point exactlybelow the ax is the centre of mass traveled a angular displacement of \(90^{\circ}=\frac{\pi}{2}\) rad \(\theta=\frac{\pi}{2} \mathrm{rad}\)

But we have \(\omega^{2}-\omega_{0}^{2}=2 \alpha \theta\)

\(\omega_{0}=0\) (Initial angular velocity) Then \(\omega=\sqrt{2 \alpha \theta}\)

$$ \begin{array}{l} =\sqrt{(2)\left(21.78 \mathrm{rad} / \mathrm{s}^{2}\right)\left(\frac{\pi}{2} \mathrm{rad}\right)} \\ =\sqrt{68.424(\mathrm{rad} / \mathrm{s})^{2}} \\ =\sqrt{8.27 \mathrm{rad} / \mathrm{s}} \end{array} $$