Question

Two equal and opposite forces are applied
tangentially to a uniform disc of mass M and radius
Ras shown in the figure. If the disc is pivoted at
its centre and free to rotate in its plane, the
angular acceleration of the disc is
2F
3 MR
MR
4F
MR
(4) Zero​

Answer 1

Answer:

Given:

2 equal and opposite forces acting tangentially on a disc , free to rotate about the centre.

Mass of disc = M

Radius of disc = R

To find:

Angular acceleration of the disc.

Concept:

The 2 tangential forces are forming a couple and providing torque to the disc. Since the disc can rotate about it's centre, it will have an angular acceleration.

Formulas used:

ζ = I × α ,

where ζ is torque, I is the moment of Inertia, α is the angular acceleration.

Calculation:

Net TORQUE = 2 × (F × R)

Applying the above formula :

2 × FR = I × α

=> 2FR = (MR²/2) × α

=> α = (4F/MR).

Answer 2

$\huge\bf{Answer:-}$

Given:

• Two equal and opposite forces are applied

tangentially to a uniform disc of mass M and radius

• If the disc is pivoted at its centre and free to rotate in its plane, the angular acceleration of the disc is 2F , 3 MR , MR , 4F , MR , (4) Zero

Hence:

• Let "M" be mass of disc.

• Let "R" be radius of disc.

Using formula:

• ζ = I × α

About formula:

• ζ = Torque.

• I = movement of interia.

• α = angular acceleration.

Explanation:

• Torque, moment, or moment of force is the rotational force.

• Just as a linear force is a push or pull, a torque can be thought of as the turning of an object.

• In three dimensions, the torque is a camouflage , For point particles, this is given by the cross product of the position vector distance vector and the force vector.

Calculation:

= ζ = 2 × ( F × R)

= 2 × FR = 1 × a

= 2 FR = ( MR² / 2 ) × a

= a = 4F / MR

.•. ( a = 4F / MR ) is the answer.