A solid sphere of mass 1 kg and radius 1 m ro | vedclass

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

Question

(D) Given: Mass $M = 1 kg$,Radius $R = 1 m$,$	heta = 30^{\circ}$,$g = 10 m s^{-2}$.Component of weight down the plane: $Mg \sin 	heta = 1 	imes 10

Vedclass · Accepted Answer

$2.86$

Vedclass · Answer

(D) Given: Mass $M = 1 kg$,Radius $R = 1 m$,$	heta = 30^{\circ}$,$g = 10 m s^{-2}$.Component of weight down the plane: $Mg \sin 	heta = 1 	imes 10 	imes \sin 30^{\circ} = 10 	imes 0.5 = 5 N$.Let $f$ be the friction force acting up the plane. The two external forces of $1 N$ each act in opposite directions relative to the center,creating a net torque.Equation of motion for translation: $Mg \sin 	heta - f - F_{ext} = Ma$. Here,the two $1 N$ forces act such that one aids motion and one opposes it,but they create a net torque. Looking at the diagram,the net force along the incline is $Mg \sin 	heta = 5 N$. The two $1 N$ forces cancel each other out in terms of net translational force.So,$5 - f = 1 	imes a \Rightarrow f = 5 - a$.Equation of motion for rotation about the center of mass $(COM)$: $	au_{net} = I \alpha$.The torque due to friction $f$ is $fR$ (counter-clockwise). The torque due to the two $1 N$ forces is $2 	imes (1 N 	imes 0.5 m) = 1 N m$ (clockwise).Thus,$fR - 1 = I \alpha$,where $I = \frac{2}{5} MR^2 = 0.4 kg m^2$ and $\alpha = \frac{a}{R} = a$.$f(1) - 1 = 0.4 a \Rightarrow f = 1 + 0.4 a$.Equating the two expressions for $f$: $5 - a = 1 + 0.4 a$.$4 = 1.4 a \Rightarrow a = \frac{4}{1.4} = \frac{40}{14} = \frac{20}{7} \approx 2.86 m s^{-2}$.

Similar Questions

$A$ solid sphere is rolling down an inclined plane without slipping. The ratio of its rotational kinetic energy to its total kinetic energy is:

What is the acceleration parallel to the surface of an inclined plane for a rolling body? Also,write the equation for the friction force acting parallel to the surface of the slope for a rolling body.

$A$ thin uniform,circular ring is rolling down an inclined plane of inclination $30^{\circ}$ without slipping. Its linear acceleration along the inclined plane will be

$A$ solid cylinder of radius $R$ and mass $M$ rolls down an inclined plane without slipping and reaches the bottom with a speed $v$. The speed would be less than $v$ if we use:

$A$ spherical ball of mass $20 \ kg$ is stationary at the top of a hill of height $100 \ m$. It rolls down a surface to the ground,then climbs up another hill of height $30 \ m$ and finally rolls down to a horizontal base at a height of $20 \ m$ above the ground. The velocity attained by the ball is

Vedclass Test Series

Exam Paper Generator

Online Exam Module