Let's analyze the problem systematically. Here is the content and what the question entails:

A 0.5 kg object rotates freely in a vertical circle at the end of a string of length 2 meters. At the top of the circle (point $P$), the tension in the string is 20 N. Assume $g = 10 \, \text{m/s}^2$.

Parts to solve: (a) Draw and label the forces acting on the object at the top of the circle.
(b) Calculate the speed of the object at point $P$.
(c) Calculate the increase in kinetic energy as the object moves from point $P$ (top) to point $Q$ (bottom).
(d) Calculate the tension in the string at the bottom point $Q$.

Let’s proceed step by step:

(a) Forces acting at point $P$:

At the top of the circle, two forces act on the object:

1. Weight ($F_g = mg$): Acts vertically downward.
2. Tension ($T$): Acts downward along the string.

The total centripetal force ($F_c$) is the net force that maintains the circular motion, given by:
$F_c = T + F_g$

(b) Speed of the object at point $P$:

From centripetal force, we have:
$F_c = \frac{mv^2}{r}$ Here:

• $m = 0.5 \, \text{kg}$,
• $r = 2 \, \text{m}$,
• $F_c = T + F_g = 20 + (0.5)(10) = 20 + 5 = 25 \, \text{N}$.

Substitute into the equation: $25 = \frac{0.5 \cdot v^2}{2}$ $v^2 = \frac{25 \cdot 2}{0.5} = 100$ $v = \sqrt{100} = 10 \, \text{m/s}$

The speed of the object at point $P$ is 10 m/s.

(c) Increase in kinetic energy from $P$ to $Q$:

The object gains potential energy as it moves downward and converts it into kinetic energy.

At the top ($P$), total energy is: $\text{Total Energy at P} = KE_P + PE_P$ At the bottom ($Q$): $\text{Total Energy at Q} = KE_Q$

The change in potential energy is: $\Delta PE = mg \cdot (2r) = (0.5)(10)(4) = 20 \, \text{J}$

The increase in kinetic energy ($\Delta KE$) is equal to the loss of potential energy: $\Delta KE = 20 \, \text{J}$

(d) Tension at the bottom point $Q$:

At the bottom, the forces contributing to centripetal force are:

1. Tension ($T_Q$): Acts upward along the string.
2. Weight ($F_g$): Acts downward.

Net force is: $F_c = T_Q - F_g$ Rewriting: $T_Q = F_c + F_g$ The centripetal force at $Q$ is: $F_c = \frac{mv_Q^2}{r}$ From conservation of energy: $KE_P + PE_P = KE_Q$ $\frac{1}{2}mv_P^2 + mg \cdot (2r) = \frac{1}{2}mv_Q^2$ Substitute known values: $\frac{1}{2}(0.5)(10^2) + (0.5)(10)(4) = \frac{1}{2}(0.5)v_Q^2$ $25 + 20 = \frac{0.25v_Q^2}{2}$ $45 = 0.25v_Q^2$ $v_Q^2 = 180 \quad \Rightarrow \quad v_Q = \sqrt{180} \approx 13.42 \, \text{m/s}$

Now calculate $T_Q$: $T_Q = \frac{(0.5)(13.42^2)}{2} + 5$ $T_Q = \frac{(0.5)(180)}{2} + 5 = 45 + 5 = 50 \, \text{N}$

The tension at the bottom is 50 N.

Summary of Results:

(a) Forces: Weight and Tension act downward at $P$.
(b) Speed at $P$: 10 m/s.
(c) Increase in KE: 20 J.
(d) Tension at $Q$: 50 N.

Would you like more detailed derivations or diagrams? Let me know!

Relative Questions:

1. What is the role of tension in maintaining circular motion in this system?
2. How does conservation of energy simplify solving for speed changes in circular motion?
3. How would the results change if the string length were increased?
4. What happens if the object’s mass is doubled?
5. How does gravitational force vary in this problem and why is it constant?

Tip:

Always separate forces acting radially and tangentially when analyzing circular motion problems.