A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination θ=30° from the horizontal. Two forces of magnitude 1 N each, parallel to the incline, act on the sphere, both at distance r=0.5 m from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is ms -2. (Take g=10 m s -2.)

Question

A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination θ=30° from the horizontal. Two forces of magnitude 1 N each, parallel to the incline, act on the sphere, both at distance r=0.5 m from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is ms -2. (Take g=10 m s -2.) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/229611707a8ef8b07ed65bd4d62f1ba1-.png" /> 5+1-1- f =1 a 5- f = a text About COM f 1-2(1(0.5))=(2/5) Mr 2 α ⇒ f -1=(2/5) a ⇒ f =1+(2/5) a 5- a =1+(2/5) a ⇒ 4=(7 a /5) ⇒ a =(20/7)=2.86 m / s 2

5+1−1−f=1a 5−f=a About COM f1−2(1(0.5))=52​Mr2α ⇒f−1=52​a ⇒f=1+52​a 5−a=1+52​a ⇒4=57a​ ⇒a=720​=2.86m/s2

$5 + 1 - 1 - f = 1 a$
$5 - f = a$
$About COM$
$f 1 - 2 (1 (0.5)) = \frac{2}{5} M r^{2} α$
$\Rightarrow f - 1 = \frac{2}{5} a$
$\Rightarrow f = 1 + \frac{2}{5} a$
$5 - a = 1 + \frac{2}{5} a$
$\Rightarrow 4 = \frac{7 a}{5}$
$\Rightarrow a = \frac{20}{7} = 2.86 m / s^{2}$