Question

4. A small object of mass M is swinging around in a vertical circle held by a massless string of length L, attached to a pivot, as shown. Given that the speed of the mass at the bottom of the circle is Vo, find the tension in the string when the mass is at an angle (theta) with respect to the vertical (see figure) Pivot M

Answer 1

The diagram of the problem is given below, Where T is the tension on the streng, Les the length of the string and I is the angle made, and also Mis the mass. If re be the velocity of the mass at the paint A then we may write from the diagram that T- Mg cose - Mr (centripetal fore) > T = Mes 4 mg case 1 Mgcose From this we can find out the unknown quantity es re and -vation law. Therefore, tension on the string. But the only that can be find out by energy conser- Energy of the mass at B = Energy of the mass at A. → 12 mv. ² + Mgl = 13 M² + Mgle-cose) = 2 M 2 2 = MV² + MgL (1-1 + lose ) => 2 Mire? - MV² + Mg L cose - re2 = vo + 29 L cose re= v vo ² tege case This will give the velocity at B as well as any point on the circumference of the circle. Therefore the tension is, T = M ( Ve + 2gh Cose) + Mg coso This is the required expression.