Question 260: Centre of Mass and Rotational Motion

Question

A particle of mass m is moving along a
trajectory given by
x = x
0
+ a cos$$\omega $$
1
t
y = y
0
+ b sin$$\omega $$
2
t
The torque, acting on the particle about the
origin, at t = 0 is :

Accepted Answer

$$\overrightarrow F  = m\overrightarrow a  = m\left[ { - a\omega _1^2\cos \omega ,t\widehat i - b\omega _2^2\sin {\omega _2}t\widehat j} ight]$$
$${\overrightarrow f _{t = 0}} =  - ma\omega _1^2\widehat i$$
$${\overrightarrow r _{t = 0}} = \left( {{X_0} + a} ight)\widehat i + y\widehat j$$
$$\overrightarrow 	au   = \overrightarrow r  	imes \overrightarrow F  = m{y_0}a\omega _1^2\widehat k$$

Answer

Zero

Answer

+my
0
a $$\omega _1^2$$$$\widehat k$$

Answer

$$ - m\left( {{x_0}b\omega _2^2 - {y_0}a\omega _1^2} \right)\widehat k$$

Answer

m (–x
0
b + y
0
a) $$\omega _1^2$$$$\widehat k$$

Centre of Mass and Rotational Motion

A particle of mass m is moving along a trajectory given by x = x 0 + a cos$\omega 1 t y = y 0 + b sin\omega $ 2 t The torque, acting on the particle about the origin, at t = 0 is :

Choose the correct answer: