A football of radius R is kept on a hole of radius r ( r

Question

A football of radius R is kept on a hole of radius r ( r < R ) made on a plank kept horizontally. One end of the plank is now lifted so that it gets tilted making an angle θ from the horizontal as shown in the figure below. The maximum value of θ so that the football does not start rolling down the plank satisfies (figure is schematic and not drawn to scale) (A)  sin θ=( r / R ) (B)  tan θ=( r / R ) (C)  sin θ=( r /2 R ) (D)  cos θ=( r /2 R )

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fc806d3b2845e37e3f19790380ca34b8-.png" /> At the verge of toppling, N 1=0 mg sin θ h = mg cos θ r and cos θ=( h / R ) So, m g sin θ h=m g (h/R) r ⇒ sin θ= r / R

$sin θ = \frac{r}{R}$

$tan θ = \frac{r}{R}$

$sin θ = \frac{r}{2 R}$

$cos θ = \frac{r}{2 R}$

At the verge of toppling, $N_{1} = 0$
$m g sin θ h = m g cos θ r$ and $cos θ = \frac{h}{R}$
So, $m g sin θ h = m g \frac{h}{R} r$
$\Rightarrow sin θ = r / R$

sinθ=Rr​

tanθ=Rr​

sinθ=2Rr​

cosθ=2Rr​

At the verge of toppling, N1​=0 mgsinθh=mgcosθr and cosθ=Rh​ So, mgsinθh=mgRh​r ⇒sinθ=r/R

$sin θ = \frac{r}{R}$

$tan θ = \frac{r}{R}$

$sin θ = \frac{r}{2 R}$

$cos θ = \frac{r}{2 R}$

At the verge of toppling, $N_{1} = 0$
$m g sin θ h = m g cos θ r$ and $cos θ = \frac{h}{R}$
So, $m g sin θ h = m g \frac{h}{R} r$
$\Rightarrow sin θ = r / R$