- Home
- Standard 11
- Physics
A uniform disc is acted by two equal forces of magnitude F. One of them, acts tangentially to the disc, while other one is acting at the central point of the disc. The friction between disc surface and ground surface is $nF$. If $r$ be the radius of the disc, then the value of $n$ would be (in $N$ )
$0$
$1.2$
$2$
$3.2$
Solution
(a)
Let $f _{ r }$ be the friction exerting between dise surface and ground surface, then the motion of disc.
we have,
From force equation
$2 F – f _{ r }= ma$
From torque equation
$\left(F+f_r\right) r=I \alpha$
$\left(F+f_r\right) r=\frac{1}{2}{m r^2}^2 \times \frac{a}{r} \ldots \ldots(2)$
here, $a$ is the linear acceleration of the disc.
$\alpha$ is the angular acceleration of the disc.
I is the moment of inertia about the centre
from equation $(1)$ and $(2)$ On solving we get
$\left( f _{ I }=0\right)$
