Question

A uniform disk with mass 36.5 kg and radius 0.220 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest,and then a constant force 35.0 N is applied tangent to the rim of the disk.

What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.390 revolution?

What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.390 revolution?

Answer 1

\(\tau=r \times F=\mid a\)

\(0.220 \times 35=36.5 \times 0.220^{2} / 2 \times a\)

\(a=7.7 / 0.8833=8.72 \mathrm{rad} / \mathrm{s}^{2}\)

after 0.390 revolution \(=0.390 \times 2 \pi=2.45\) rad

\(\omega^{2}=2 a \theta\)

\(\omega=\sqrt{(2} \times 8.72 \times 2.45)=6.54 \mathrm{rad} / \mathrm{s}\)

\(v_{t}=\omega r=6.54 \times 0.220=1.44 \mathrm{~m} / \mathrm{s}\)

there will be centripiatl acceleration and tangential accel.

\(a_{t}=a X r=8.72 \times 0.22=1.92 \mathrm{~m} / \mathrm{s}^{2} \quad\) tangentially to disk.

\(a_{c}=v_{t}^{2} / r=1.44^{2} / 0.22=9.43 \mathrm{~m} / \mathrm{s}^{2}\) towards centre of disk

both are at 90 degree angle

so resultant acce. \(a=\sqrt{1} .92^{2}+9.43^{2}=92.53 \mathrm{~m} / \mathrm{s}^{2}\)