Place a uniform meter scale horizontally on y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $N_1$ and $N_2$ be the normal forces on the left and right fingers respectively. For a uniform meter scale of mass $M$ and length $100 \ cm$,t

Vedclass · Accepted Answer

$25.60$

Vedclass · Answer

(A) Let $N_1$ and $N_2$ be the normal forces on the left and right fingers respectively. For a uniform meter scale of mass $M$ and length $100 \ cm$,the center of mass is at $50 \ cm$. Initially,the fingers are at $0 \ cm$ and $90 \ cm$. The distance of the center of mass from the left finger is $50 \ cm$ and from the right finger is $40 \ cm$.For rotational equilibrium about the center of mass: $N_1(50) = N_2(40) \implies 5N_1 = 4N_2$.Also,$N_1 + N_2 = Mg$. Substituting $N_2 = 1.25N_1$,we get $2.25N_1 = Mg \implies N_1 = \frac{4}{9}Mg$ and $N_2 = \frac{5}{9}Mg$.When the left finger slips,it experiences kinetic friction $f_{k1} = \mu_k N_1$ and the right finger experiences static friction $f_{s2} \le \mu_s N_2$. When the left finger stops and the right starts slipping,the right finger experiences kinetic friction $f_{k2} = \mu_k N_2$ and the left finger experiences static friction $f_{s1} = \mu_s N_1$.At the point where the right finger stops and the left finger starts slipping,the torque about the center of mass is zero: $N_1 x_L = N_2 x_R$.Also,the condition for the transition is $f_{s1} = f_{k2} \implies \mu_s N_1 = \mu_k N_2$.Given $\mu_s = 0.40$ and $\mu_k = 0.32$,we have $0.40 N_1 = 0.32 N_2 \implies N_1 = 0.8 N_2 = \frac{4}{5} N_2$.Substituting this into the torque equation: $(\frac{4}{5} N_2) x_L = N_2 x_R \implies x_R = 0.8 x_L$.From the previous stage where the left finger stopped,$N_1 x_L = N_2(40)$ with $4N_1 = 5N_2$ (i.e.,$N_1 = 1.25 N_2$),we found $x_L = 32 \ cm$.Thus,$x_R = 0.8 	imes 32 = 25.6 \ cm$.

Similar Questions

If an explosion occurs in a projectile,which force acts on its centre of mass?

Provide an example from daily life where motion is controlled or initiated by an external force.

$A$ $100 \, kg$ gun fires a ball of $1 \, kg$ horizontally from a cliff of height $500 \, m$. It falls on the ground at a distance of $400 \, m$ from the bottom of the cliff. Find the recoil velocity of the gun. (Acceleration due to gravity $g = 10 \, m/s^2$) (in $, m/s$)

$A$ bouncing ball of mass $200 \,g$ falls from a height of $5 \,m$ onto a horizontal ground. After every impact with the ground, the velocity of the ball decreases by $\frac{1}{2}$ times. The total momentum the ball imparts to the ground after $3$ impacts is (Let $g=10 \,m/s^2$):

$A$ force of $mg$ is applied to block $B$ to keep the system in equilibrium. Find the value of $T_1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module