Answer to Question #249635 in Molecular Physics | Thermodynamics for Kelani

Question

The uniform thin rod in the figure below has mass M = 2.00 kg and length L = 3.73 m and is free to rotate on a frictionless pin. At the instant the rod is released from rest in the horizontal position, find the magnitude of the rod's angular acceleration, the tangential acceleration of the rod's center of mass, and the tangential acceleration of the rod's free end.

(a)

the rod's angular acceleration (in rad/s²)

___ rad/s²

(b)

the tangential acceleration of the rod's center of mass (in m/s²)

___ m/s²

(c)

the tangential acceleration of the rod's free end (in m/s²)

___ m/s²

Answer 1

Solution;

Given;

M=2.0kg

L=3.73m

Take;

$\alpha$ as the the Angular acceleration.

T as torque.

I as moment of inertia.

a)

From Newton's second law;

$T=I\alpha$

If ;

T=$\frac{WL}{2}$

I=$\frac{ML^2}{3}$

Hence;

$\frac{WL}{2}=\frac{ML^2}{3}\alpha$

Hence the Angular acceleration is given as;

$\alpha=\frac{3g}{2L}$ =$\frac{3×9.82}{2×3.73}=3.95$

$\alpha=3.95$ rad/s²

b)

Tangential acceleration of rods center of mass is;

$a_t=r\alpha$

r is radius=$\frac L2$

$a_t=\frac L2×\frac{3g}{2L}$

$a_t=\frac{3.73}2×3.95$

$a_t=7.367m/s^2$

(c)

Tangential acceleration of rods free end;

$v=L\alpha$

$v=3.73×3.95$

$v=14.73m/s^2$