Question #249635

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

1
Expert's answer
2021-10-17T16:53:20-0400

Solution;

Given;

M=2.0kg

L=3.73m

Take;

α\alpha as the the Angular acceleration.

T as torque.

I as moment of inertia.

a)

From Newton's second law;

T=IαT=I\alpha

If ;

T=WL2\frac{WL}{2}

I=ML23\frac{ML^2}{3}

Hence;

WL2=ML23α\frac{WL}{2}=\frac{ML^2}{3}\alpha

Hence the Angular acceleration is given as;

α=3g2L\alpha=\frac{3g}{2L} =3×9.822×3.73=3.95\frac{3×9.82}{2×3.73}=3.95

α=3.95\alpha=3.95 rad/s2

b)

Tangential acceleration of rods center of mass is;

at=rαa_t=r\alpha

r is radius=L2\frac L2

at=L2×3g2La_t=\frac L2×\frac{3g}{2L}

at=3.732×3.95a_t=\frac{3.73}2×3.95

at=7.367m/s2a_t=7.367m/s^2

(c)

Tangential acceleration of rods free end;

v=Lαv=L\alpha

v=3.73×3.95v=3.73×3.95

v=14.73m/s2v=14.73m/s^2


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Comments

Stephen
04.04.23, 23:37

Excellent explanation. Thanks!

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