Question

A tangential force of 20 N is applied on a cylinder of mass
4 kg and moment of inertia 0.02 kg m about its own axis ir
the cylinder rolls without slipping, then linear acceleration
of its centre of mass will be
(A) 6,7m
(b) 10 m/s
(c) 3.3m's
(d) None of these​

Answer 1

Given:

Force = 20 N

Mass = 4 kg

Moment of Inertia = 0.02 kg m²

To find:

The linear acceleration of its center of mass.

Solution:

• We know that a = rα

• Also, as per the question, 20 + f = 4a

• And, 20 x 0.1 - f x 0.1 = 0.02α

• Upon solving the above equations, we get,

        20+f = 4α                    eq(1)

        2-0.1f = 0.02 a/r

        20-f = (0.2 a) / 0.1         eq(2)

• On solving eq(1) and eq(2), we get,

        a = 6.7 m/s²

• Thus, the linear acceleration of its centre of mass will be 6.7 m/s².

Answer 2

Given :

Tangential Force = F = 20 N

The mass of cylinder = M = 4 kg

Moment of Inertia = I = 0.02 kg-m

To Find :

The linear acceleration of its center of mass

Solution :

Let The linear acceleration = a  m/s²

Moment of Inertia of Cylinder = I = $\dfrac{Mr^{2} }{2}$

Or, 2 I = M × r²

Or, 2 × 0.02 = 4 × r²

or, r = 0.1 m

∵  T = I a

So, ( F - f ) r = $\dfrac{Mr^{2} }{2}$ × a

Or, 2 ( F - f ) r = Mr² × a

∴   a  = $\dfrac{2 ( F - f ) r}{Mr^{2} }$            ...........1

Again

∵  acceleration of its center of mass = $\dfrac{ ( F + f ) }{M}$            ..........2

From eq1  and eq 2

$\dfrac{2 ( F - f ) r}{Mr^{2} }$ = $\dfrac{ ( F + f ) }{M}$

Or,  2 ( F - f ) = r ( F + f )

Or, 2  ( F - f ) = 0.1 × ( F + f )

Or, 20 F - 20 f = F  + f

Or,  19 F = 21 f

i.e       F = $\dfrac{21}{19}$ f

Or, f = $\dfrac{19F}{21}$

∴  acceleration of its center of mass = $\dfrac{ ( F + \dfrac{19F}{21} ) }{M}$

                                                             = $\dfrac{40 F}{21 M}$

                                                             = $\dfrac{40\times 20}{21\times 4}$

                                                             = 9.52 m/s² = 10  m/s²

Hence, The acceleration of its center of mass is  10  m/s²   Answer