Question

Best,

I am stuck with a physics problem. The question relates to the domain of the dynamics of the point particle. The assignment is the following: A smooth block with a mass of 10 kg is shot in half a loop. The contact between the block and the loop is perfectly smooth (no friction). The radius of the looping is 5 m (see figure). There is no air resistance. The block is separated from the looping in the indicated position C.

The question that is asked is to determine the radius of curvature ρ of the track (oblique throw) at 1.00 seconds after coming off in C. Can you help me?

5 30° 50° m-10kg

Answer 1

First understand qualitatively. After loosing contact at C, the block follows a parabolic path (like a projectile). Using the equation of the parabola you just have to find the radius of curvature.

First find the speed at point C.

Loosing contact translates to normal force becoming 0.

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

where g=10, R=5 m and $\theta=30\degree$. Now set N=0,

$v=\sqrt {gR\sin\theta}$

$v=5\textup{ m/s}$

Projectile motion begins.

Launch speed v=5 m/s at an angle of $\alpha=60\degree$.

Define the equation of parabola using parameter t (time):-

$x=vt\cos\alpha,\ y=vt\sin\alpha-\frac{1}{2}gt^2$

The origin of this coordinate system is at C and axis x points to left.

Formula for radius of curvature

$\rho=\frac{\left[1+\left ( dy/dx \right )^2\right]^{3/2}}{|d^2y/dx^2|}$

$\frac{dx}{dt}=v\cos\alpha,\ \frac{dy}{dt}=v\sin\alpha-gt$

$\Rightarrow\frac{dy}{dx}=\frac{v\sin\alpha-gt}{v\cos\alpha}$

at t=1

$\frac{dy}{dx}=\sqrt{3}-\frac{10}{5\cdot 1/2}=\sqrt{3}-4$

and

$\frac{d^2y}{dx^2}=-\frac{g}{v\cos\alpha}\frac{dt}{dx}$

$\Rightarrow \frac{d^2y}{dx^2}=-\frac{g}{v^2\cos^2\alpha}=-1.6$

On putting the values we get

$\rho=9.52\textup{ m}$