JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 10)
A particle of mass m is moving along a
trajectory given by
x = x0 + a cos$$\omega $$1t
y = y0 + b sin$$\omega $$2t
The torque, acting on the particle about the origin, at t = 0 is :
x = x0 + a cos$$\omega $$1t
y = y0 + b sin$$\omega $$2t
The torque, acting on the particle about the origin, at t = 0 is :
Zero
+my0a $$\omega _1^2$$$$\widehat k$$
$$ - m\left( {{x_0}b\omega _2^2 - {y_0}a\omega _1^2} \right)\widehat k$$
m (–x0b + y0a) $$\omega _1^2$$$$\widehat k$$
Explanation
$$\overrightarrow F = m\overrightarrow a = m\left[ { - a\omega _1^2\cos \omega ,t\widehat i - b\omega _2^2\sin {\omega _2}t\widehat j} \right]$$
$${\overrightarrow f _{t = 0}} = - ma\omega _1^2\widehat i$$
$${\overrightarrow r _{t = 0}} = \left( {{X_0} + a} \right)\widehat i + y\widehat j$$
$$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = m{y_0}a\omega _1^2\widehat k$$
$${\overrightarrow f _{t = 0}} = - ma\omega _1^2\widehat i$$
$${\overrightarrow r _{t = 0}} = \left( {{X_0} + a} \right)\widehat i + y\widehat j$$
$$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = m{y_0}a\omega _1^2\widehat k$$
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