The Change Friction Makes In The Complexity Of A Problem.

First, we make the force diagram at the instant when then angle made by line joining center of circle, and block with horizontal is $\theta$ . The force diagram would be as shown. Let the speed of the block at this instant be $v$ . Clearly, there would be two components of acceleration , centripetal and tangential.

We know that centripetal acceleration is given by $\frac{v^2}{R}$ , hence we use Newton's law to get:

$N - mg \sin \theta = \frac{mv^2}{R}$

$\Rightarrow N = mg \sin \theta + \frac{mv^2}{R}$

We know that $f = \mu_{k} N$ , $\Rightarrow f = \mu_{k} (mg \sin \theta + \frac{mv^2}{R})$ .

Now, tangential acceleration = $\frac{dv}{dt} = \frac{v dv}{v dt} = \frac{v dv}{R d \theta} = \frac{d(v^2)}{2 R d \theta}$ , hence, using Newton's law,

$mg \cos \theta - \mu_{k} (mg \sin \theta + \frac{mv^2}{R}) = m \frac{d(v^2)}{2 R d \theta}$

Rearrange and divide by $m$ to get:

$\frac{d(v^2)}{d \theta} + 2 \mu_{k} v^2 = 2 Rg(\cos \theta - \mu_{k} \sin \theta)$

Clearly, this is a Linear Differential equation in $v^2$ and $\theta$ .

Integrating factor = $\displaystyle e^{\int 2 \mu_{k} d \theta} = e^{2 \mu_{k} \theta}$

Hence the solution is given by:

$v^2 e^{2 \mu_{k} \theta} = 2Rg \displaystyle \int_{0}^{\theta} e^{2 \mu_{k} \theta} \big(\cos \theta - \mu_{k} \sin \theta\big) d \theta$

The right hand side can be easily evaluated , if you have already seen this post. Here, clearly, $a = 2 \mu_{k} ; b = 1$ .

Solving, we get:

$\displaystyle \boxed{ v^2 = \frac{2 Rg}{4 \mu_{k}^2 + 1} \bigg(3 \mu_{k} (\cos \theta - e^{- 2 \mu_{k} \theta}) + (1 - 2 \mu_{k}^2) \sin \theta \bigg)}$

In our case, $\mu_{k} = 0.5$ , and we have to find the value of $v$ at bottom (say $v_{b})$ , hence, $\theta = \frac{\pi}{2}$ . Substituting these values,

$v_{b} = \sqrt{\frac{Rg}{2} (1 - 3e^{- \frac {\pi}{2}})}$ .

Now, using the formula for the range of a projectile given horizontal velocity from a given height, we get:

$\displaystyle x = \sqrt{\frac{Rg}{2} (1 - 3e^{- \frac {\pi}{2}}) \times \frac{2h}{g}} = \boxed{\sqrt{Rh(1 - 3e^{- \frac{\pi}{2}})}}$

Substitute values to get : $x \approx 0.89 m = 89 cm$

First of All, Beautiful question. Congratulation to the one who formed it. I thoroughly enjoyed solving it.

Now to the solution.

A schematic Diagram can be found here

As is apparent from the diagram the frictional force is $\mu (MgSin \theta +Mv^{2}/R)$

Using Newton's second Law,

$M \frac{dv}{dt}=MgCos \theta- \mu (MgSin \theta +Mv^{2}/R)$

Dividing throughout by $M$ ,

$\frac{dv}{dt}=gCos \theta- \mu (gSin \theta +v^{2}/R)$

And by rules of rotational dynamics,

$v=R \frac{d\theta}{dt}$

Eliminating $dt$ from the equations we get,

$v \frac{dv}{d\theta}=gRCos \theta- \mu gRSin \theta -\mu v^{2}$

$v \frac{dv}{d\theta}$ can be written as $\frac{1}{2} \frac{dv^{2}}{d\theta}$

which gives us the linear differential equation,

$\frac{dv^{2}}{d\theta}+ 2\mu v^{2}=2gR(Cos \theta- \mu Sin \theta)$

Solving the equation for $\theta$ from $0$ to $\pi /2$ , we will get the required value of $v$

For its solution, the Integrating factor would be $e^{2 \mu \theta}$

Therefore,

$v^{2}e^{2 \mu \theta}\bigg|_{0}^{\pi/2}=2gR\int_0^{\pi/2} e^{2 \mu \theta}(Cos \theta- \mu Sin \theta)d \theta$

The Left hand side can be easily expanded as $v^{2}e^{ \mu \pi}$

To solve the Integral on the Right hand side, we designate the integral as $I$ and apply Integration by parts twice

$I=\int_0^{\pi/2} e^{2 \mu \theta}(Cos \theta- \mu Sin \theta)d \theta$

Applying by parts,

$I=\frac{1}{2 \mu} (e^{2 \mu \theta}(Cos \theta- \mu Sin \theta)\bigg|_{0}^{\pi/2} +\int_0^{\pi/2} e^{2 \mu \theta}(Sin \theta + \mu Cos \theta)d \theta)$

This can be called as $Equation 1$

The first term can be evaluated as $e^{2 \mu \theta}(Cos \theta- \mu Sin \theta)\bigg|_{0}^{\pi/2}= -\mu e^{ \mu \pi} -1$

The integral can be evaluated as

$\int_0^{\pi/2} e^{2 \mu \theta}(Sin \theta + \mu Cos \theta)d \theta= \frac{1}{2 \mu} (e^{2 \mu \theta}(Sin \theta + \mu Cos \theta)\bigg|_{0}^{\pi/2} - \int_0^{\pi/2} e^{2 \mu \theta}(Cos \theta - \mu Sin \theta)d \theta)$

The first term is $e^{2 \mu \theta}(Sin \theta + \mu Cos \theta)\bigg|_{0}^{\pi/2}= e^{ \mu \pi}- \mu$

and the second term is $I$ itself.

Replacing in $Equation 1$ ,

$I=\frac{1}{2 \mu} ( -\mu e^{ \mu \pi} -1+ \frac{1}{2 \mu} (e^{ \mu \pi}- \mu -I))$

On solving for $I$ , we get

$I=(e^{\pi \mu}(\frac{1}{2 \mu}-\mu)-\frac{3}{2})/(2 \mu +\frac{1}{2 \mu})$

By putting the value of $\mu = 0.5$ ,

$I=\frac{1}{4}(e^{\pi/2}-3)$

Putting its value in the original equation,

$v^{2}e^{ \mu \pi}=gR\frac{1}{2}(e^{\pi/2}-3)$

Therefore, the speed with which it is released from the ramp is

$v=\sqrt{gR \frac{1-3e^{-\pi/2}}{2}}$

The range covered $x=v \sqrt{\frac{2h}{g}}$

Putting the value of $v$ ,

$x= \sqrt{hR(1-3e^{- \pi/2})}$

Putting the value of $h=105cm$ and $R=200cm$ ,

We get $x=88.9021185906$

Approximating to the nearest integer,

$x=\boxed{89}$