91Ó°ÊÓ

To solve this problem, we will apply Newton's second law in the radial direction, as there is no net force in the tangential direction. The centripetal force on the rock is provided by the tension T in the rope. According to the centripetal force formula, we have: \(F_c = m\frac{v^2}{r}\), with r being the Lenght \(L\) of the rope. So, the equation for the tension in the rope is: \(T = m\frac{v^2}{L}\).

To find the tension considering the gravitational force, we need to draw a free-body diagram for the rock. There are two forces acting on the rock: the tension in the rope and the weight of the rock (mg). The tension acts along the rope, while gravity acts downward. We can resolve the tension force into two components: horizontal and vertical. The horizontal component of tension provides the centripetal force required for circular motion, while the vertical component balances the weight of the rock. According to the problem, the angle between the rope and the horizontal is \(\theta\). Therefore, the horizontal component of the tension, \(T_x = T\cos(\theta)\) and the vertical component, \(T_y = T\sin(\theta)\). The centripetal force acting on the rock is provided by the horizontal component of tension, so \(T_x = m\frac{v^2}{L}\). Also, we have gravitational force acting downwards, and vertical component acts upwards, so \(T_y = mg\). Now we can solve for T using the components: \(T\cos(\theta) = m \frac{v^2}{L}\) and \(T\sin(\theta) = mg\). Squaring and summing these two equations, we get: \((T\cos(\theta))^2 + (T\sin(\theta))^2 = (m\frac{v^2}{L})^2 + (mg)^2\). Simplifying, we get: \(T^2(\cos^2(\theta) + \sin^2(\theta)) = m^2(\frac{v^4}{L^2} + g^2v^2)\). Since \(\cos^2(\theta)+ \sin^2(\theta) = 1\), the equation for tension becomes: \(T^2 = m^2(\frac{v^4}{L^2} + g^2v^2)\). Taking the square root of both sides, we find the tension in the rope: \(T = m(\sqrt{\frac{v^4}{L^2} + g^2v^2})\).