Q 28 8 A disc rolls down a plane of length L and inclined at - Physics - Communication Systems

Question

Q 28

8     A   disc   rolls   down   a
  plane   of   length   L   and  
  inclined   at   angle   θ ,  
without   slipping   Its   velocity   on  
reaching   the   bottom   will   be   : -
1   4 gL   sin   θ 3      
                 
                 
                 
                2   2
gL   sin   θ 3          
                 
                 
              3   10 gL
  sin   θ 7          
                 
                 
              4   4 gL
  sin   θ

Pintu B. · Accepted Answer

Dear Student ,
<img alt="" src=
"https://img-nm%20mnimgs%20com/img/study_content/content_ck_images/images/rolling(1)%20png"
style=
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The Equation of linear motion of centre of mass is given
by
∑Fx= max⇒mgsinθ-f=ma

The Equation of rotational motion of centre of mass is given
by
∑Γx=Ια ; where Γ is total torque, I is momen
t of inertia about c m
 and αis the angular acceleration of cylinder⇒fr=12mr2α⇒fr=12mr2ar   , as we know a=αr
⇒f=12ma      
(1)putting this value in above equation, we getmgsinθ-12ma=ma⇒a=23gsinθ

Let the speed be v, so from the formula
v2=u2+2as⇒v2=2×23gsinθ×l=43glsinθThus v=43glsinθ

Regards

Q 28 8 . A disc rolls down a plane of length L and inclined at angle θ , without slipping . Its velocity on reaching the bottom will be : - 1 4 gL sin θ 3 2 2 gL sin θ 3 3 10 gL sin θ 7 4 4 gL sin θ

Q 28

$8 . A disc rolls down a plane of length L and inclined at angle θ, without slipping . Its velocity on reaching the bottom will be : - (1) \sqrt{\frac{4 gL \sin θ}{3}} (2) \sqrt{\frac{2 gL \sin θ}{3}} (3) \sqrt{\frac{10 gL \sin θ}{7}} (4) \sqrt{4 gL \sin θ}$