Pathfinder for Olympiad and JEE Advanced Physics
LEVELJEE Advanced
Visualized Solution
The Sigma Insight: Kinematics of Rotational Motion
The Geometry of Winding
Solving the Tape Roll Problem
Imagine a tape being wound onto a spool. We know the initial radius of the empty spool, the final radius of the full roll, the total length of the tape, and the total time it takes to wind it. Our goal is to find out how much tape is wound at a specific intermediate time.
At first glance, this looks like a terrifying calculus problem involving spirals and changing velocities. But what if I told you there is a beautiful geometric shortcut?
The Area-Length Trick
Here is a brilliant trick. Instead of dealing with complex spirals, think about the cross-sectional area of the tape roll.
The face area of the wound tape is simply the total length of the tape multiplied by its thickness. This means the area is directly proportional to the length.
This simple realization transforms a messy kinematics problem into an elegant geometry puzzle.
Kinematics of the Radius
Since the motor rotates the spool at a constant angular velocity , the number of turns increases linearly with time.
And because each turn adds a constant thickness , the radius of the roll also increases linearly from the initial radius to the final radius .
This linear relationship is the key to unlocking the intermediate state of the tape roll.
The Master Equation
Now, let's look at the tape wound up to time . Its area will be proportional to the length wound up to that time.
By taking the ratio of this intermediate area to the total area, the unknown thickness perfectly cancels out!
This gives us our master equation. The fraction of the length wound is exactly equal to the fraction of the area filled.
Final Calculation
Let's plug in the numbers to find the radius at . We substitute the initial and final radii, and the given times into our linear radius equation.
Finally, we substitute this radius back into our master equation. We square the radii to find the areas.
Three hundred over five hundred and twenty-five simplifies beautifully to . Multiplying this by the total length of gives us exactly . And that is our final answer!
Similar Questions
JEE Main 2021, 20 July Shift-II
LEVELJEE Main
A body rotating with an angular speed of is uniformly accelerated to in . The number of rotations made in the process is …… .
JEE Main 2021, 17 March Shift-I
LEVELJEE Main
The angular speed of truck wheel is increased from to in . The number of revolutions by the truck engine during this time is ……… . (Assuming the acceleration to be uniform).
LEVELJEE Advanced
A pulley of radius is rotated about its axis by a force (where, is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is , then the number of rotations made by the pulley before its direction of motion is reserved, is
(A)
more than 3 but less than 6
(B)
more than 6 but less than 9
(C)
more than 9
(D)
less than 3
JEE Advanced 2012
LEVELJEE Advanced
Consider a disc rotating in the horizontal plane with a constant angular speed about its centre . The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles and are simultaneously projected at an angle towards . The velocity of projection is in the - plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed rotation, (ii) their range is less than half the disc radius, and (iii) remains constant throughout. Then
(A)
lands in the shaded region and in the unshaded region
(B)
lands in the unshaded region and in the shaded region
(C)
both and land in the unshaded region
(D)
both and land in the shaded region
JEE Advanced 2012
LEVELJEE Main
Comprehension Passage
The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed , the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the Z-axis, and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points and ). Both these motions have the same angular speed in this case.
Now, consider two similar systems as shown in the figure : Case (a) the disc with its face vertical and parallel to x-z plane; Case (b) the disc with its face making an angle of with x-y plane and its horizontal diameter parallel to X-axis. In both the cases, the disc is welded at point , and the systems are rotated with constant angular speed about the Z-axis.
Question 1:
Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct?
(A)
It is for both the cases
(B)
It is for case (a); and for case (b)
(C)
It is for case (a); and for case (b)
(D)
It is for both the cases
Question 2:
Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct?
(A)
It is vertical for both the cases (a) and (b)
(B)
It is vertical for case (a); and is at to the x-z plane and lies in the plane of the disc for case (b)
(C)
It is horizontal for case (a); and is at to the x-z plane and is normal to the plane of the disc for case (b)
(D)
It is vertical for case (a); and is at to the x-z plane and is normal to the plane of the disc for case (b)
JEE Advanced 2017
LEVELJEE Advanced
A rigid uniform bar of length is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vertical is . Which of the following statements about its motion is/are correct?
* Multiple Correct Options
(A)
Instantaneous torque about the point in contact with the floor is proportional to
(B)
The trajectory of the point is parabola
(C)
The mid-point of the bar will fall vertically downward
(D)
When the bar makes an angle with the vertical, the displacement of its mid-point from the initial position is proportional to