A man pushes a cylinder of mass `m_1` with the help of a plank of mass `m_2` as shown in figure. There in no slipping at any contact. The horizontal component of the force applied by the man is F.
(a) the acceleration fo the plank and the center of mass of the cylinder, and
(b) the magnitudes and direction of frictional force at contact points.
A man pushes a cylinder of mass `m_1` with the help of a plank of mass `m_2` as shown in figure. There in no slipping at any contact. The horizontal component of the force applied by the man is F.
(a) the acceleration fo the plank and the center of mass of the cylinder, and
(b) the magnitudes and direction of frictional force at contact points.
(a) the acceleration fo the plank and the center of mass of the cylinder, and
(b) the magnitudes and direction of frictional force at contact points.
Text Solution
Verified by Experts
The correct Answer is:
A, B, C, D
The man applies a force F in the horizontal direction on the
plank as shown. Therefore the point of contact of the plank
with the cylinder will try to move towards right. Therefore
the friction force F will act towards left on the plank. To
each and every action there is equal and opposite reaction.
Therefore a frictional force f will act on the top of the cylinder
towards right.
Direction of f' : A force f is acting on the cylinder. This force
is trying to move the point of contact P towards right by an
acceleration
`a_(cm) = (f)/(M_1) acting towards right.
At the same time, the force f is trying to rotate the cylinder
abuout its centre of mass.
`fxxR = Ixxalpha`
`rArr alpha = (fxxR)/(I) = (fxxR)/(1)/(2)M_1R^2 = (2f)/(M_1R)` in clockwise direction.
`:. alpha_(cm) + alphaR=(f)/(M_1) - (2f)/(M_1R) xxR = -(f)/(M_1,` i.e. towards left.
Therefore, the point of constact of the cylinder with the
ground move towards left. Hence frictin force acts towards
right on the cylinder.
NOTE: You ca assume any direction of friction at the point
of contact and solve the problem. If the value of friction
comes out to be positivem, our assumed direction is correct
otherwise the direction of friction is opposite. The above
activity is done so that if only the direction of friction is
asked, an approach may be developed.
Applying Newton's law on plank, we get
`F-f = m_2a_2 .....(i)`
Also, `a_2 = 2a_1 ...(ii)`
Because `a_2` is the accelration of topmost point of cylinder
and there is no slipping.
Applying Newton's no cylinder
`M_1a_1 = f+f ....(iii)`
The torque equation for the cylinder is
`fxx R - fxxR= Ialpha = (1)/(2)M_1 R^2xx((a_1)/(R ))`
`:. I = (1)/(2) M_1R^2 and Ralpha = a_1]`
`:. (f-f) R = (1)/(2) M_1Ra_1 rArr f+f=(1)/(2)M_1a_1 .....(4)`
Solving equation (iii) and (iv), we get
`f = (3)/(4)M_1a_1....(5)`
and `f = (1)/(4)M_1 a_1 ...(6)
From (i) and (iii)
`F - f=2m_2 a_1 rArr F- (3)/(4)M_1a_1 = 2m_2a_1`
`:. a_1 = (4f)/(3M_1+8m_2) :. a_2 = (8F)/(3M_1+8m_2)`
From (v) and (vi)
`f = (3)/(4)M_1xx(4f)/(3M_1+8m_2) = (3FM_1)/(3M_1+8m_2)`
and `f'=(1)/(4)M_1xxa_1 = (FM_1)/(3M_1+ 8m_2)`
plank as shown. Therefore the point of contact of the plank
with the cylinder will try to move towards right. Therefore
the friction force F will act towards left on the plank. To
each and every action there is equal and opposite reaction.
Therefore a frictional force f will act on the top of the cylinder
towards right.
Direction of f' : A force f is acting on the cylinder. This force
is trying to move the point of contact P towards right by an
acceleration
`a_(cm) = (f)/(M_1) acting towards right.
At the same time, the force f is trying to rotate the cylinder
abuout its centre of mass.
`fxxR = Ixxalpha`
`rArr alpha = (fxxR)/(I) = (fxxR)/(1)/(2)M_1R^2 = (2f)/(M_1R)` in clockwise direction.
`:. alpha_(cm) + alphaR=(f)/(M_1) - (2f)/(M_1R) xxR = -(f)/(M_1,` i.e. towards left.
Therefore, the point of constact of the cylinder with the
ground move towards left. Hence frictin force acts towards
right on the cylinder.
NOTE: You ca assume any direction of friction at the point
of contact and solve the problem. If the value of friction
comes out to be positivem, our assumed direction is correct
otherwise the direction of friction is opposite. The above
activity is done so that if only the direction of friction is
asked, an approach may be developed.
Applying Newton's law on plank, we get
`F-f = m_2a_2 .....(i)`
Also, `a_2 = 2a_1 ...(ii)`
Because `a_2` is the accelration of topmost point of cylinder
and there is no slipping.
Applying Newton's no cylinder
`M_1a_1 = f+f ....(iii)`
The torque equation for the cylinder is
`fxx R - fxxR= Ialpha = (1)/(2)M_1 R^2xx((a_1)/(R ))`
`:. I = (1)/(2) M_1R^2 and Ralpha = a_1]`
`:. (f-f) R = (1)/(2) M_1Ra_1 rArr f+f=(1)/(2)M_1a_1 .....(4)`
Solving equation (iii) and (iv), we get
`f = (3)/(4)M_1a_1....(5)`
and `f = (1)/(4)M_1 a_1 ...(6)
From (i) and (iii)
`F - f=2m_2 a_1 rArr F- (3)/(4)M_1a_1 = 2m_2a_1`
`:. a_1 = (4f)/(3M_1+8m_2) :. a_2 = (8F)/(3M_1+8m_2)`
From (v) and (vi)
`f = (3)/(4)M_1xx(4f)/(3M_1+8m_2) = (3FM_1)/(3M_1+8m_2)`
and `f'=(1)/(4)M_1xxa_1 = (FM_1)/(3M_1+ 8m_2)`
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