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                                                                    Rotational	motion	problems	and	solutions
  Question	-1	Find	the	Moment	of	Inertia	of	a	sphere	with	axis	tangent	to	it?	Solution	The	moment	of	inertia	of	the	sphere	about	the	axis	passing	through	the	center	us	$I_C=\frac	{2}{5}MR^2$	Using	Parallel	axis	theorem,	Moment	of	inertia	through	the	tangent	is	given	by	$I_T	=I_C	+	MR^2$	or	$I_T=	{7}{5}MR^2$	Question	-2	A	car	accelerates
  uniformly	from	rest	and	reaches	a	Speed	of	22.0	m/s	in	9.00	s.	If	the	radius	of	a	tire	is	29	cm,	find	the	number	of	revolutions	the	tire	makes	during	this	motion,	assuming	that	no	slipping	occurs.	Solution	From	$v=u+at$	$a=	\frac	{v}{t}$	Now	Distance	covered	$s=	\frac	{1}{2}at^2	=	\frac{1}{2}v	t$	Number	of	revolutions=	$\frac	{s}{2	\pi	r}$
  Substituting	the	values	Number	of	revolutions=54.3	revolutions	Question	3	A	cylinder	of	Radius	20	cm	is	rolling	along	the	surface	with	constant	speed	of	80	cm/sec	a.	What	is	the	rotational	speed	of	the	cylinder	about	its	axis?	b.	What	is	the	magnitude	and	direction	of	the	acceleration	of	the	points	on	it	surface.	c.	what	is	the	velocity	of	the	point	at	the
  top	of	the	cylinder	at	any	moment	d.	what	is	the	velocity	of	the	point	of	the	cylinder	which	is	in	contact	with	the	floor	at	any	moment	Solution	a.	Rotational	speed	$\omega	=	\frac	{v}{R}	=	\frac	{80}{20}	=	4	rad/s$	b.Since	there	is	no	acceleration	for	cylinder	,	$\alpha	=	0$	and	we	have	just	radial	acceleration	$a_r=	\omega	^2	r=	3.2	m/s^2$	c.
  Velocity	at	the	top	will	be	given	as	$v_T	=	v	+	r	\omega	=1.6	m/s	$	d.	Velocity	at	the	top	will	be	given	as	$v_f=0$	Question	4	The	angular	momentum	of	a	particle	moving	in	a	circular	orbit	with	a	constant	speed	remains	conserved	about	a.	Any	point	of	the	circumference	of	the	circle	b.	any	point	inside	the	circle	c.any	point	outside	the	circle	d.	the
  centre	of	the	circle	Solution	Answer	is	(d)	Question	5	Two	discs	of	Moment	of	inertia	$I_1$	and	$I_2$	about	their	respective	axis	rotating	with	angular	velocities	$\omega	_1$	and	$\omega	_2$	respectively	are	brought	in	face	to	face	with	their	axis	of	rotation	coincident.	The	angular	velocity	of	the	composite	disc	will	be	a.$\frac	{I_1	\omega_1	+	I_2
  \omega_2}{I_1	+	I_2}$	b.$\frac	{I_1	\omega_1	-	I_2	\omega_2}{I_1	-	I_2}$	c.	$\frac	{I_2	\omega_1	+	I_1	\omega_2}{I_1	+	I_2}$	d.	$\frac	{I_2	\omega_1	-	I_1	\omega_2}{I_1	+	I_2}$	Solution	Answer	is	(a)	Angular	momentum	is	conserved	as	no	external	torque	$I_1	\omega_1	+	I_2	\omega_2	=	(I_1	+	I_2)	\omega$	or	$\omega	=\frac	{I_1	\omega_1	+
  I_2	\omega_2}{I_1	+	I_2}$	Question	6	\frac	{I_1	\omega_1	+	I_2	\omega_2}{I_1	+	I_2}	Question	-6	A	cylinder	rolls	up	an	inclined	plane,	reaches	some	height	and	then	rolls	down	(without	slipping	throughout	these	motions).	The	directions	of	the	frictional	force	acting	on	the	cylinder	are,	a.up	the	incline	while	ascending	and	down	the	incline	while
  descending.	b.up	the	incline	while	ascending	as	well	as	descending.	c.down	the	incline	while	ascending	and	up	the	incline	while	descending.	d.	down	the	incline	while	ascending	as	well	as	descending.	Solution	Suppose	the	inclination	of	the	plane	is	$\alpha$.	The	force	acting	on	the	cylinder	are	weight	mg	and	normal	reaction	N.	In	the	absence	of
  frictional	force,	the	component	$mgsin	\alpha$	will	give	an	acceleration	down	the	incline	and	hence	contact	point	will	have	a	tendency	to	move	downward.	To	oppose	this	tendency,	the	frictional	force	will	act	upward.	This	is	independent	of	whether	cylinder	is	moving	up	or	down.	Hence	answer	is	(b)	Question	-7	A	disc	has	a	speed	of	1200	rpm	and	it	is
  made	to	slow	down	at	a	uniform	rate	of	4	rad/s2.Calculate	the	number	of	revolution	it	makes	before	coming	to	rest?	Solution	f=1200	rpm	=	20	rotation	per	sec	Initial	angular	velocity	of	the	disc	($\omega_0$)	=$	2	\pi	f=	2	\pi	\times	20	=40	\pi	$	rad/s	Final	angular	velocity	of	the	disc	($\omega_f$)=0	Angular	acceleration	(($\alpha$)	=	-4	rad/s2	Now
  $\omega_f=	\omega	_0	+	\alpha	t$	or	$t=	\frac	{\omega_0}{	\alpha	}	=	10	\pi$	sec	Now	$\theta	=	\omega	_0	t	+	\frac	{1}{2}	\alpha	t^2$	$\theta=200	\pi^2$	rad	Therefore,	Number	of	revolutions=	$\frac	{200	\pi^2}{2	\pi}	=	314$	Question	8	A	circular	disc	of	mass	m	and	radius	r	is	rolling	on	a	smooth	horizontal	surface	with	a	constant	speed	v.
  Calculate	the	Kinetic	energy	of	the	disc	Solution	Kinetic	Energy	of	the	rolling	disc	is	given	by	$K	=	\frac	{1}{2}mv^2	+	\frac	{1}{2}	I\omega^2$	Now	$I	=	\frac	{1}{2}mr^2$	and	$	\omega	=	\frac	{v}{r}$	So,	$K	=	\frac	{1}{2}mv^2	+	\frac	{1}{2}	(\frac	{1}{2}mr^2)	(\frac	{v}{r})^2$	$=	\frac	{3}{4}mv^2	$	Question	9	A	small	mass	m	is
  attached	to	a	massless	string	whose	other	end	is	fixed	at	P	as	shown	in	the	figure.	The	mass	is	undergoing	circular	motion	is	the	x-y	plane	with	centre	at	O	and	constant	angular	speed	?.	If	the	angular	momentum	of	the	system,	calculated	about	O	and	P	are	denoted	by	$\vec{L_O}$	and	$\vec{L_P}$	respectively,	then	(A)	$\vec{L_O}$	and	$\vec{L_P}$
  do	not	vary	with	time.	(B)	$\vec{L_O}$	varies	with	time	while	$\vec{L_P}$	remains	constant.	(C)	$\vec{L_O}$	remains	constant	while	$\vec{L_P}$	varies	with	time.	(D)	$\vec{L_O}$	and	$\vec{L_P}$	both	vary	with	time	Solution	$\vec{L_0}$	is	constant	both	in	magnitude	and	direction	$\vec{L_P}$	is	constant	in	magnitude	,	but	direction	varies
  Question	10	A	bug	of	mass	m	sits	on	the	edge	of	the	circular	disc	of	Radius	R	and	rotating	with	angular	speed	$\omega$.	How	large	must	be	the	coefficient	of	friction	between	bug	and	disc	be	if	the	bug	is	not	to	slip	off?	Solution	For	the	bug	not	to	slip	off,	the	frictional	force	$f=\mu	mg	$	must	supplies	the	centripetal	force	So,	$\mu	mg	=	m	\omega^2
  R$	or	$\mu	=	\frac	{\omega^2	R}{g}$	link	to	this	page	by	copying	the	following	textRotational	motion	problems	with	solutions	Also	Read	When	you	try	to	solve	problems	of	Physics	in	general	and	of	rotational	motion	in	particular,	it	is	important	to	follow	a	certain	order.	Try	to	be	organized	when	you	solve	these	problems,	and	you	will	see	how	it	gives
  good	results.	It	is	worth	spending	a	bit	of	time	on	the	analysis	of	a	problem	before	tackling	it.	When	solving	a	rotational	motion	problem,	follow	these	steps:	Knowledge	is	free,	but	servers	are	not.	Please	consider	supporting	us	by	disabling	your	ad	blocker	on	YouPhysics.	Thanks!	Read	carefully	the	problem	statement.	Draw	a	diagram	of	the	physical
  situation	described	in	the	problem.	Write	down	the	givens	in	the	problem	statement.	Identify	the	elements	that	constitute	the	system	in	motion:	whether	it	is	a	single	body	or	if	on	the	contrary	you	must	take	into	account	several	objects.	This	step	is	important	because	it	will	allow	you	to	distinguish	between	the	internal	and	external	forces	that	act	on
  the	system.	Draw	the	external	forces	that	act	on	the	system.	Remember	that	forces	are	physical	interactions.	Therefore,	if	a	system	is	supported	on	another	system,	a	normal	force	will	act	on	it;	if	it	is	close	to	the	surface	of	the	Earth,	the	weight	will	act	on	it,	and	so	on	with	the	rest	of	the	interactions	that	the	system	experiences.	In	this	link	you	can	see
  how	the	forces	exerted	by	the	different	types	of	support	are	represented.	Add	a	Cartesian	coordinate	system	to	the	drawing,	indicating	which	is	the	positive	direction	of	each	of	the	axes.	you	should	also	indicate	which	point	you	are	going	to	use	as	an	origin	to	calculate	the	torque	of	the	external	forces	acting	on	the	system.	If	the	center	of	mass	of	the
  system	moves,	write	Newton’s	second	law	for	it	as	a	vectorial	equation,	including	in	the	vectorial	sum	for	the	different	external	forces	you	have	identified.	Project	the	resulting	equation	on	the	Cartesian	axes.	Take	into	account	the	sign	of	the	different	projections.	If	the	system	rotates,	write	Newton’s	second	law	for	rotation	in	a	vectorial	form.
  Calculate	the	torques	of	the	different	external	forces	with	respect	to	the	point	you	have	chosen	as	origin.	Project	the	resulting	equation	on	the	Cartesian	axes.	Take	into	account	the	sign	of	the	different	projections.	Solve	the	system	of	equations	that	you	have	obtained	after	projecting	the	different	equations	on	the	axes.	In	this	way	you	will	be	able	to
  determine	the	magnitudes	asked	in	the	problem.	Do	not	forget	to	include	the	units	in	your	results.	Review	the	problem	and	check	that	the	results	you	have	obtained	make	sense.	On	the	following	pages	you	will	find	some	rotational	motion	problems	with	solutions.	Try	to	do	them	before	looking	at	the	solution.	The	post	Rotational	motion	problems	and
  solutions	appeared	first	on	YouPhysics	Rotational	Motion	Exam1	and		Problem	Solutions	1.	An	object,	attached	to	a	0,5m	string,	does	4	rotation	in	one	second.	Find	a)	Period	b)	Tangential	velocity	c)	Angular	velocity	of	the	object.		a)	If	the	object	does	4	rotation	in	one	second,	its	frequency	becomes;	f=4s-1	T=1/f=1/4s	b)	Tangential	velocity	of	the
  object;	V=2.π.f.r	V=2.3.4.0,5	V=12m/s	c)	Angular	velocity	of	the	object	ω=2.π.f=2.3.4=24radian/s	2.	Find	the	relation	between	tangential	and	angular	velocities	of	points	X,	Y	and	Z.	X	and	Y	rotate	together,	so	if	X	does	one	rotation	then	Y	also	does	one	rotation.	On	the	contrary,	if	Y	does	one	revolution,	Z	does	two	revolutions.	Angular	velocities	of	the
  X,	Y	and	Z	are;	ωX=ωY=ωZ/2	3.	An	object	hanged	on	a	rope	L=0,5m,	does	rotational	motion.	If	the	angle	between	rope	and	vertical	is	370,	find	the	tangential	velocity	of	the	object.	(g=10m/s2,	cos370=0,8,	sin370=0,6)	Free	body	diagram	of	system	is	given	below;	Horizontal	component	of	tension	on	the	rope	makes	object	rotate.		TX=mV2/r,	TY=m.g
  Radius	of	the	motion	path	is;	r=L.sin370=0,5.0,6=0,3m	tan370=TX/TY	3/4=mV2/r/m.g	3/4=V2/g.r	V=3/2m/s	4.	An	object	having	mass	m	does	rotational	motion.	Its	angular	velocity	is	ω	and	radius	of	motion	path	is	r.	Find	kinetic	energy	of	the	object	in	terms	of	r,	ω,	and	m.	EK=1/2m.V2	V=ω.r	EK=1/2m(ω.r)2	EK=mω2.r2/2	5.	Stone	having	mass	0,5kg
  rotates	in	horizontal.	It	is	hanged	on	1m	rope.	If	the	tension	on	the	rope	is	80	N,	find	the	frequency	of	the	motion.	Fnet=80N=m.ω2.r	80=m.4.π2.f2.r	80=0,5.4.32.f2.1	f=2s-1	Tags:	Torque	1.	A	beam	140	cm	in	length.	There	are	three	forces	acts	on	the	beam,	F1	=	20	N,	F2	=	10	N,	and	F3	=	40	N	with	direction	and	position	as	shown	in	the	figure
  below.	What	is	the	torque	causes	the	beam	rotates	about	the	center	of	mass	of	the	beam?	Known	:	The	center	of	mass	located	at	the	center	of	the	beam.	Length	of	beam	(l)	=	140	cm	=	1.4	meters	Force	1	(F1)	=	20	N,	the	lever	arm	1	(l1)	=	70	cm	=	0.7	meters	Force	2	(F2)	=	10	N,	the	lever	arm	2	(l2)	=	100	cm	–	70	cm	=	30	cm	=	0.3	meters	Force	3
  (F3)	=	40	N,	the	lever	arm	3	(l3)	=	70	cm	=	0.7	meters	Wanted	:	The	magnitude	of	torque	Solution	:	The	torque	1	rotates	beam	clockwise,	so	assigned	a	negative	sign	to	the	torque	1.	τ1	=	F1	l1	=	(20	N)(0.7	m)	=	-14	N	m	The	torque	2	rotates	beam	counterclockwise,	so	assigned	a	positive	sign	to	the	torque	2.	τ2	=	F2	l2	=	(10	N)(0.3	m)	=	3	N	m	The
  torque	3	rotates	beam	clockwise,	so	assigned	a	positive	sign	to	the	torque	3.	τ3	=	F3	l3	=	(40	N)(0.7	m)	=	-28	N	m	The	net	torque	:	Στ	=	-14	Nm	+	3	Nm	–	28	Nm	=	–	42	Nm	+	3	Nm	=	-39	Nm	The	magnitude	of	the	torque	is	39	N	m.	The	direction	of	rotation	of	the	beam	clockwise,	so	assigned	a	negative	sign.	Read	:		Graphical	of	linear	motion	–
  problems	and	solutions	2.	What	is	the	net	torque	acts	on	the	beam	The	axis	of	rotation	at	point	D.	(sin	53o	=	0.8)	Known	:	The	axis	of	rotation	at	point	D	F1	=	10	N	and	l1	=	r1	sin	θ	=	(40	cm)(sin	53o)	=	(0.4	m)(0.8)	=	0.32	meters	F2	=	10√2	N	and	l2	=	r2	sin	θ	=	(20	cm)(sin	45o)	=	(0.2	m)(0.5√2)	=	0.1√2	meters	F3	=	20	N	and	l3	=	r1	sin	θ	=	(10	cm)
  (sin	90o)	=	(0.1	m)(1)	=	0.1	meters	Wanted	:	The	net	torque	Solution	:	τ1	=	F1	l1	=	(10	N)(0.32	m)	=	3.2	Nm	(The	torque	1	rotates	beam	counterclockwise	so	we	assign	positive	sign	to	the	torque	1)	τ2	=	F2	l2	=	(10√2	N)(	0.1√2	m)	=	-2	Nm	(The	torque	2	rotates	beam	clockwise	so	we	assign	negative	sign	to	the	torque	2)	τ3	=	F2	l2	=	(20	N)(0.1	m)	=
  2	Nm	(The	torque	3	rotates	beam	counterclockwise	so	we	assign	positive	sign	to	the	torque	3)	The	net	torque	:	Στ	=	τ1	–	τ1	+	τ3	Στ	=	3.2	Nm	–	2	Nm	+	2	Nm	Στ	=	3.2	Nm	Read	:		Radioactivity	–	problems	and	solutions	3.	What	is	the	net	torque	if	the	axis	of	rotation	at	point	D.	(sin	53o	=	0.8)	Known	:	The	axis	of	rotation	at	point	D.	Distance	between
  F1	and	the	axis	of	rotation	(rAD)	=	40	cm	=	0.4	m	Distance	between	F2	and	the	axis	of	rotation	(rBD)	=	20	cm	=	0.2	m	Distance	between	F3	and	the	axis	of	rotation	(rCD)	=	10	cm	=	0.1	m	F1	=	10	Newton	F2	=	10√2	Newton	F3	=	20	Newton	Sin	53o	=	0.8	Wanted	:	The	net	torque	Solution	:	The	moment	of	the	force	1	Στ1	=	(F1)(rAD	sin	53o)	=	(10	N)
  (0.4	m)(0.8)	=	3.2	N.m	(The	torque	1	rotates	beam	counterclockwise	so	we	assign	positive	sign	to	the	torque	1)	The	moment	of	the	force	2	Στ2	=	(F2)(rBD	sin	45o)	=	(10√2	N)(0.2	m)(0.5√2)	=	-2	N.m	(The	torque	2	rotates	beam	clockwise	so	we	assign	negative	sign	to	the	torque	2)	The	moment	of	the	force	3	Στ3	=	(F3)(rCD	sin	90o)	=	(20	N)(0.1	m)(1)
  =	2	N.m	(The	torque	2	rotates	beam	counterclockwise	so	we	assign	positive	sign	to	the	torque	3)	The	net	torque	:	Στ	=	Στ1	+	Στ2	+	Στ3	Στ	=	3.2	–	2	+	2	Στ	=	3.2	Newton	meter	Read	:		Determine	the	position	of	an	object	in	projectile	motion	The	moment	of	inertia	4.	Length	of	wire	=	12	m,	l1	=	4	m.	Ignore	wire’s	mass.	What	is	the	moment	of	inertia
  of	the	system.	Known	:	Mass	of	A	(mA)	=	0.2	kg	Mass	of	B	(mB)	=	0.6	kg	Distance	between	A	and	the	axis	of	rotation	(rA)	=	4	meters	Distance	between	B	and	the	axis	of	rotation	(rB)	=	12	–	4	=	8	meters	Wanted	:	The	moment	of	inertia	of	the	system	Solution	:	The	moment	of	inertia	of	A	IA	=	(mA)(rA2)	=	(0.2)(4)2	=	(0.2)(16)	=	3.2	kg	m2	The	moment
  of	inertia	of	B	IB	=	(mB)(rB2)	=	(0.6)(8)2	=	(0.6)(64)	=	38.4	kg	m2	The	moment	of	inertia	of	the	system	:	I	=	IA	+	IB	=	3.2	+	38.4	=	41.6	kg	m2	Read	:		Pressure	of	solids	–	problems	and	solutions	Rotational	dynamics	5.	A	6-N	force	is	applied	to	a	cord	wrapped	around	a	pulley	of	mass	M	=	5	kg	and	radius	R	=	20	cm.	What	is	the	angular	acceleration	of
  the	pulley.	The	pulley	is	a	uniform	solid	cylinder.	Known	:	Force	(F)	=	6	Newton	Mass	(M)	=	5	kg	Radius	(R)	=	20	cm	=	20/100	m	=	0.2	m	Wanted	:	Angular	acceleration	(α)	Solution	:	The	moment	of	the	force	:	τ	=	F	R	=	(6	Newton)(0.2	meters)	=	1.2	N	m	The	moment	of	inertia	for	solid	cylinder	:	I	=	1/2	M	R2	I	=	1/2	(5	kg)(0.2	m)2	I	=	1/2	(5	kg)(0.04
  m2)	I	=	1/2	(0.2)	I	=	0.1	kg	m2.	The	angular	acceleration	:	τ	=	I	α	α	=	τ	/	I	=	1.2	/	0.1	=	12	rad	s-2	Read	:		Optical	instrument	contact	lenses	–	problems	and	solutions	6.	A	block	of	mass	=	4	kg	hanging	from	a	cord	wrapped	around	a	pulley	of	mass	=	8	kg	and	radius	R	=	10	cm.	Acceleration	due	to	gravity	is	10	ms-2	.	What	is	the	linear	acceleration	of
  the	block?	The	pulley	is	a	uniform	solid	cylinder.	Known	:	Mass	of	pulley	(m)	=	8	kg	Radius	of	pulley	(r)	=	10	cm	=	0.1	m	Mass	of	block	(m)	=	4	kg	Acceleration	due	to	gravity	(g)	=	10	m/s2	Weight	(w)	=	m	g	=	(4	kg)(10	m/s2)	=	40	kg	m/s2	=	40	Newton	Wanted	:	The	free	fall	acceleration	of	the	block	Solution	:	The	moment	of	inertia	of	the	solid
  cylinder	:	I	=	1/2	M	R2	=	1/2	(8	kg)(0.1	m)2	=	(4	kg)(0.01	m2)	=	0.04	kg	m2	The	moment	of	the	force	:	τ	=	F	r	=	(40	N)(0.1	m)	=	4	Nm	The	angular	acceleration	:	Στ	=	I	α	4	=	0.04	α	α	=	4	/	0.04	=	100	The	linear	acceleration	:	a	=	r	α	=	(0.1)(100)	=	10	m/s2	7.	A	block	with	mass	of	m	hanging	from	a	cord	wrapped	around	a	pulley.	If	the	free	fall
  acceleration	of	the	block	is	a	m/s2,	what	is	the	moment	of	inertia	of	the	pulley..	Known	:	weight	=	w	=	m	g	Lever	arm	=	R	The	angular	acceleration	=	α	The	free	fall	acceleration	of	the	block	=	a	ms-2	Wanted:	The	moment	of	inertia	of	the	pulley	(I)	Solution	:	The	connection	between	the	linear	acceleration	and	the	angular	acceleration	:	a	=	R	α	α	=	a	/
  R	The	moment	of	inertia	:	τ	=	I	α	I	=	τ	:	α	=	τ	:	a	/	R	=	τ	(R	/	a)	=	τ	R	a-1	Read	:		Springs	in	series	and	parallel	–	problems	and	solutions	The	angular	momentum	8.	A	0.2-gram	particle	moves	in	a	circle	at	a	constant	speed	of	10	m/s.	The	radius	of	the	circle	is	3	cm.	What	is	the	angular	momentum	of	the	particle?	Known	:	Mass	of	particle	(m)	=	0.2	gram
  =	2	x	10-4	kg	Angular	speed	(ω)	=	10	rad	s-1	Radius	(r)	=	3	cm	=	3	x	10-2	meters	Wanted	:	The	angular	momentum	of	the	particle	Solution	:	The	equation	of	the	angular	momentum	:	L	=	I	ω	I	=	the	angular	momentum,	I	=	the	moment	of	inertia,	ω	=	the	angular	speed	The	moment	of	inertia	(for	particle)	:	I	=	m	r2	=	(2	x	10-4	)(3	x	10-2)2	=	(2	x	10-4	)
  (9	x	10-4)	=	18	x	10-8	The	angular	momentum	:	L	=	I	ω	=	(18	x	10-8)(10	rad	s-1)	=	18	x	10-7	kg	m2	s-1
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