Question

1. A rod of length l and radius r is joined to a rod of length l/2 and radius r/2 of same material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of A degrees. The twist angle at the joint will be (ans-8A/9)

2.A uniform wire of length l and radius r is twisted by angle A. if modulus of rigidity of wire is B then elastic potential energy stored in wire is?(ans-pie*B*r^4*A^2/4l)

3.Two wires A and B are of same length and of same material having radii r1 and r2 respectively. Their one end is fixed with a rigid support and at the other end equal twisting couple is applied. Then ratio of the angle of twist at the end of A and the angle of twist at the end B will be? ans-r1^2/r2^2

Answer 1

Answer on Question #50913-Physics-Mechanics-Kinematic-Dynamics

1. A rod of length $l$ and radius $r$ is joined to a rod of length $\frac{l}{2}$ and radius $\frac{r}{2}$ of same material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of $A$ degrees. The twist angle at the joint will be

Solution

$$\tau = \tau^ {\prime}$$ $$\frac {\pi B r ^ {4}}{2 l} (A - A ^ {\prime}) = \frac {\pi B \left(\frac {r}{2}\right) ^ {4}}{2 \left(\frac {l}{2}\right)} A ^ {\prime}.$$ $$A - A ^ {\prime} = \frac {A ^ {\prime}}{8} \rightarrow A ^ {\prime} = \frac {8}{9} A.$$

2. A uniform wire of length $l$ and radius $r$ is twisted by angle $A$ . If modulus of rigidity of wire is $B$ then elastic potential energy stored in wire is?

Solution

An elastic potential energy stored in wire is

$$u = \frac {1}{2} \tau \cdot A = \frac {1}{2} C A ^ {2}.$$

Torque $C$ produced per unit twist of a wire of length $l$ and radius $r$ is given by

$$C = \frac {\tau}{A} = \frac {\pi B r ^ {4}}{2 l}.$$

Thus

$$u = \frac {1}{2} \frac {\pi B r ^ {4}}{2 l} A ^ {2} = \frac {\pi B r ^ {4} A ^ {2}}{4 l}.$$

3. Two wires $A$ and $B$ are of same length and of same material having radii $r_1$ and $r_2$ respectively. Their one end is fixed with a rigid support and at the other end equal twisting couple is applied. Then ratio of the angle of twist at the end of $A$ and the angle of twist at the end $B$ will be?

Solution

$$\tau = \frac {\pi B r ^ {4}}{2 l} \theta .$$

In the given problem:

$$r ^ {4} \theta = const.$$

Thus

$$\frac {\theta_ {A}}{\theta_ {B}} = \frac {r _ {2} ^ {4}}{r _ {1} ^ {4}}.$$

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