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प्रश्न
A uniform disc of radius R, is resting on a table on its rim.The coefficient of friction between disc and table is µ (Figure). Now the disc is pulled with a force F as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?

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उत्तर
Let the acceleration of the centre of mass of the disc be ‘a’, then
`Ma = F - f` ......(1)
The angular acceleration of the disc is `α = a/R`. (if there is no sliding).
Then `(1/2 MR^2)α = Rf` ......(2)

⇒ `Ma = 2f`
Thus, `f = F/3`. Since there is no sliding,
⇒ `f ≤ µmg`
⇒ `F ≤ 3µMg`.
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संबंधित प्रश्न
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Using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.
Read each statement below carefully, and state, with reasons, if it is true or false;
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Read each statement below carefully, and state, with reasons, if it is true or false;
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Read each statement below carefully, and state, with reasons, if it is true or false;
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Answer in Brief:
A rigid object is rolling down an inclined plane derive the expression for the acceleration along the track and the speed after falling through a certain vertical distance.
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If x = at + bt2, where x is the distance travelled by the body in kilometers while t is the time in seconds, then the unit of b is ______.
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(Force constant of the spring = 36 N/m)
