Animated Solution for Physics - Rotational Motion: A particle of mass M=0.2 kg is initially at rest in the xy-plane at a point (x=−l,y=−h), where l=10 m and h=1 m. The particle is accelerated at time t=0 with a constant acceleration a=10 m/s2 along the positive x-direction. Its angular momentum and torque with respect to the origin, in SI units, are represented by L and τ, respectively. i^,j^ and k^ are unit vectors along the positive x, y and z-directions, respectively. If k^=i^×j^ then which of the following statement(s) is(are) correct ?
Select Answer:
* Multiple Correct
Visualized Solution
Visualizing the Setup
Initial position: A(−10,−1)
Acceleration: a=10i^ m/s2
Kinematics Equation
S=ut+21at2
Here, initial velocity u=0
Substituting Values
Displacement S=10−(−10)=20 m
20=0+21(10)t2
Calculating Time
20=5t2
t2=4⟹t=2 s
Torque Formula
τ=r×F
Force and Position Vectors
F=ma=0.2(10i^)=2i^ N
r=xi^−1j^
Calculating Torque
τ=(xi^−j^)×(2i^)
τ=−2(j^×i^)=2k^ N⋅m
Angular Momentum Formula
L=r×p=r×(mv)
Velocity at Point B
v=at=(10i^)(2)=20i^ m/s
rB=10i^−j^
Calculating Angular Momentum
L=(10i^−j^)×(0.2×20i^)
L=(10i^−j^)×4i^
L=−4(j^×i^)=4k^ kg⋅m2/s
Conclusion
Correct Options: (A), (B), (C)
00:00 / 00:00
The Sigma Insight: Torque and Angular Momentum
Solution Diagram
Imagine you are observing a particle in the vast expanse of the xy-plane. It sits quietly at a specific coordinate, waiting for a force to act upon it. Suddenly, a constant acceleration jolts it into motion. This problem is a beautiful symphony of 1D kinematics, vector cross products, and rotational dynamics. Let's break down the journey of this particle step by step.
Visualizing the Particle's Journey
The particle starts at point A, which has coordinates (−l,−h). Given the values, this is (−10,−1). The acceleration is purely in the positive x-direction, a=10i^ m/s2. Because there is no initial velocity and no acceleration in the y-direction, the particle is constrained to move along the horizontal line y=−1.
Our first task is to determine when it reaches point B, which has coordinates (l,−h) or (10,−1). The total displacement along the x-axis is the final position minus the initial position: S=10−(−10)=20 m.
The Kinematics of the Motion
Since the acceleration is constant, we can confidently deploy the second equation of motion:
S=ut+21at2
Substituting our known values (S=20, u=0, a=10):
20=0+21(10)t2
20=5t2⟹t2=4
Solving this yields t=2 s. This perfectly matches statement (A), confirming that the particle arrives at point B exactly at t=2 seconds.
Unveiling the Torque
Next, we need to investigate the torque τ acting on the particle about the origin. Torque is defined as the cross product of the position vector r and the force vector F:
τ=r×F
First, let's find the force. According to Newton's Second Law, F=ma. With a mass of 0.2 kg, the force is:
F=0.2(10i^)=2i^ N
At any general point during its motion, the particle's position vector is r=xi^−1j^. Now, let's compute the cross product:
τ=(xi^−j^)×(2i^)
Remember your cross product rules: i^×i^=0 and j^×i^=−k^.
τ=−2(j^×i^)=−2(−k^)=2k^ N⋅m
Notice something fascinating here? The torque does not depend on the x-coordinate! It is a constant 2k^ throughout the entire motion. Therefore, when the particle passes through (l,−h), the torque is indeed 2k^. This makes statement (B) correct and statement (D) incorrect.
Calculating the Angular Momentum
Finally, let's evaluate the angular momentum L at point B. Angular momentum is the cross product of the position vector and the linear momentum p:
L=r×p=r×(mv)
We need the velocity at t=2 s. Using the first equation of motion v=u+at:
v=0+(10i^)(2)=20i^ m/s
The position vector at point B is rB=10i^−j^. Let's plug everything into the angular momentum equation:
L=(10i^−j^)×(0.2×20i^)
L=(10i^−j^)×4i^
Again, the i^×i^ term vanishes, leaving us with:
L=−4(j^×i^)=−4(−k^)=4k^ kg⋅m2/s
This confirms that statement (C) is also correct. By systematically applying fundamental principles, we have unraveled the entire physical state of the particle!
