Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Physics

Mechanics

Easy

Question

Two particles each of mass m are attached at the ends of a massless rod of length L which is on a smooth table at rest. An impulse J is applied to one particle along a direction perpendicular to the length of the rod. Immediately after that, angular velocity of the rod would be

$\frac{J}{m L}$
$\frac{2 J}{m L}$
$\frac{J}{2 m L}$
$\frac{2 J}{3 m L}$

The correct answer is: $fraction numerator J over denominator m L end fraction$

Angular impulse = change in angular momentum
$table attributes columnalign left columnspacing 1em rowspacing 4 pt end attributes row cell not stretchy rightwards double arrow fraction numerator J L over denominator 2 end fraction equals 2 m open parentheses L over 2 close parentheses squared omega end cell row cell fraction numerator J L over denominator 2 end fraction equals fraction numerator m L squared omega over denominator 2 end fraction not stretchy rightwards double arrow omega equals fraction numerator J over denominator m L end fraction end cell row cell J equals 2 m v subscript c not stretchy rightwards double arrow v subscript c equals fraction numerator J over denominator 2 m end fraction end cell end table$

Related Questions to study

A thin uniform rod of mass m and length L is force to rotate about its upper and when it is at rest, it receives an impulse J at its lowest point, normal to its length. immediately, after impact

A thin uniform rod of mass m and length L is force to rotate about its upper and when it is at rest, it receives an impulse J at its lowest point, normal to its length. immediately, after impact

PhysicsMechanics

$text A particular of mass end text m equals 5 text is moving with a uniform speed end text v equals 3 square root of 2 text in the XOY plane along the line end text Y equals X plus 4 T h e space m a g n i t u d e space o f space t h e space a n g u l a r space m o m e n t u m space o f space t h e space p a r t i c l e space a b o u t space t h e space o r i g i n space i s$

$text A particular of mass end text m equals 5 text is moving with a uniform speed end text v equals 3 square root of 2 text in the XOY plane along the line end text Y equals X plus 4 T h e space m a g n i t u d e space o f space t h e space a n g u l a r space m o m e n t u m space o f space t h e space p a r t i c l e space a b o u t space t h e space o r i g i n space i s$

PhysicsMechanics

A stone is whirled along a circle with a constant angular velocity and its angular momentum is L. If the radius of its circular path is halved while keeping the angular velocity same, its new angular momentum would be

A stone is whirled along a circle with a constant angular velocity and its angular momentum is L. If the radius of its circular path is halved while keeping the angular velocity same, its new angular momentum would be

PhysicsMechanics

parallel

A uniform rod of mass m and length 2l is rotating about an axis passing through its centre and perpendicular to its length. If is its angular velocity, its angular momentum about that axis is

A uniform rod of mass m and length 2l is rotating about an axis passing through its centre and perpendicular to its length. If is its angular velocity, its angular momentum about that axis is

PhysicsMechanics

$text A uniform circular disc of mass end text straight M text and radius end text straight R text rolls without slipping on a horizontal surface end text text If the velocity of its centre is end text v subscript 0 comma text then the total angular momentum of the disc about a fixed end text text point end text P text at a height end text 3 straight R divided by 2 text about the centre end text straight C text . end text$

$text A uniform circular disc of mass end text straight M text and radius end text straight R text rolls without slipping on a horizontal surface end text text If the velocity of its centre is end text v subscript 0 comma text then the total angular momentum of the disc about a fixed end text text point end text P text at a height end text 3 straight R divided by 2 text about the centre end text straight C text . end text$

PhysicsMechanics

A disc of mass m and radius R moves in the X-Y plane as shown in figure. The angular momentum of the disc about the origin O at the instant shown is

A disc of mass m and radius R moves in the X-Y plane as shown in figure. The angular momentum of the disc about the origin O at the instant shown is

PhysicsMechanics

parallel

A cube of mass ‘m’ and side ‘a’ moving along with a constant speed as shown in figure. The magnitude of angular momentum of cube about z-axis would be

A cube of mass ‘m’ and side ‘a’ moving along with a constant speed as shown in figure. The magnitude of angular momentum of cube about z-axis would be

PhysicsMechanics

$text Consider a body, shown in the figure, consisting of two identical balls, each of mass M end text text connected by a light rigid rod. If an impulse end text J equals M V text is imparted to the body at one of its ends, what would be end text i t s space a n g u l a r space m o t i o n$

$text Consider a body, shown in the figure, consisting of two identical balls, each of mass M end text text connected by a light rigid rod. If an impulse end text J equals M V text is imparted to the body at one of its ends, what would be end text i t s space a n g u l a r space m o t i o n$

PhysicsMechanics

$text A shell of mass "m' is projected with velocity end text v text at an angle end text 60 to the power of ring operator text to the horizontal. When it reaches the end text text maximum height, its angular momentum with respect to point of projection is end text$

$text A shell of mass "m' is projected with velocity end text v text at an angle end text 60 to the power of ring operator text to the horizontal. When it reaches the end text text maximum height, its angular momentum with respect to point of projection is end text$

PhysicsMechanics

parallel

Angular momentum of a solid sphere about its diameter is L. If it rotates about its tangent with the same angular velocity as that in the first case, now its angular momentum will be

Angular momentum of a solid sphere about its diameter is L. If it rotates about its tangent with the same angular velocity as that in the first case, now its angular momentum will be

PhysicsMechanics

$text A constant torque acting on a uniform circular disc changes its angular momentum from end text L text to end text fraction numerator 4 L over denominator 3 end fraction text in end text 2 text seconds. Then magnitude of the torque applied is end text$

$text A constant torque acting on a uniform circular disc changes its angular momentum from end text L text to end text fraction numerator 4 L over denominator 3 end fraction text in end text 2 text seconds. Then magnitude of the torque applied is end text$

PhysicsMechanics

$space A space t h i n space minus w a l l e d space p i p e space r o l l s space a l o n g space t h e space f l o o r. space W h a t space i s space t h e space r a t i o space o f space i t s space t r a n s l a t i o n a l space k i n e t i c space e n e r g y space t o space i t s space r o t a t i o n a l space k i n e t i c space e n e r g y space a b o u t space t h e space c e n t r a l space a x i s space p a r a l l e l space t o space i t s space l e n g t h space ?$

$space A space t h i n space minus w a l l e d space p i p e space r o l l s space a l o n g space t h e space f l o o r. space W h a t space i s space t h e space r a t i o space o f space i t s space t r a n s l a t i o n a l space k i n e t i c space e n e r g y space t o space i t s space r o t a t i o n a l space k i n e t i c space e n e r g y space a b o u t space t h e space c e n t r a l space a x i s space p a r a l l e l space t o space i t s space l e n g t h space ?$

PhysicsMechanics

parallel

$text A copper ball of mass end text straight m equals 1 kg text with a radius of end text straight r equals 10 cm text rotates with angular velocity end text omega equals 2 rad divided by straight s text about an axis passing through its centre. The work should be performed to increases the angular velocity of rotation of the end text text ball two fold is end text$

$text A copper ball of mass end text straight m equals 1 kg text with a radius of end text straight r equals 10 cm text rotates with angular velocity end text omega equals 2 rad divided by straight s text about an axis passing through its centre. The work should be performed to increases the angular velocity of rotation of the end text text ball two fold is end text$

PhysicsMechanics

A solid uniform disk of mass m and radius R is pivoted about a horizontal axis through its centre and a small body of mass m is attached to the rim of the disk. If the disk is released from rest with the small body, initially at the same level as the centre, the angular velocity when the small body reaches the lowest point of the disk is

A solid uniform disk of mass m and radius R is pivoted about a horizontal axis through its centre and a small body of mass m is attached to the rim of the disk. If the disk is released from rest with the small body, initially at the same level as the centre, the angular velocity when the small body reaches the lowest point of the disk is

PhysicsMechanics

$text A composite rod comprising two rods of mass end text straight m text and end text 2 straight m text and each of length end text l equals 1 straight m text as shown in the end text text figure. Assume end text omega equals 10 rad divided by straight s text and end text straight m equals 1 kg comma text the end text KE text of the rotating composite rod is end text$

$text A composite rod comprising two rods of mass end text straight m text and end text 2 straight m text and each of length end text l equals 1 straight m text as shown in the end text text figure. Assume end text omega equals 10 rad divided by straight s text and end text straight m equals 1 kg comma text the end text KE text of the rotating composite rod is end text$

PhysicsMechanics

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.