302 - Mechanics part 1 Questions Answers
if a particle is moving on a circular path with constant speed then the angle between the direction of acceleration and its position vector with respect to centre of circle will be?
since speed is constant so only centripetal acc. will be present and it will be parallel to position vector starting from centre of the circle.
the reduced mass of two particle having masses m and 2m is 2m/m.how????.....
formula for finding reduced mass will be [m1m2/m1+m2]
the moment of inertia of a thin uniform rod about an axis through its end and perpendicular to its length is ML²/3 . The moment of inertia of the rod about an axis parallel to given axis and at a distance L from the center of the rod is
for an axis parellel to the axis of rod and at distance l, rod will act as a set of equidistance particles. so answer will be Ml2
answer given by sarika is not completely correct
A wheel starting from rest is uniformly accelerated at 2 radsec^2 for 20s.It is allowed to rotate uniformly for next 10s and finaly brought to rest in next 20s.Find total angle rotated by wheel ( in radian).
ω = 0 + 2(20) = 40 (according to rotational equation ω = ω0 + αt)
and θ = 0 + 1/2(2)(20)2 = 400 radian
in middle uniform part angle rotated = 40*10 = 400 radian
and since acceleration in first part and retardation in last part are equal so angle rotated in last part = 400 radian
total = 1200 radian
A body rotating with uniform angular acceleration covers 100 π (radian) in the first 5 s after the start. Its angular speed at the end of 5 s ( in radian/s) is
initial angular velocity = 0
so use θ = 1/2 αt2
and ω = αt
if A + B + C = 0 then A x B is
It is not given that in which terms we have to get answer so many answers are possible
A possible solution
A + B + C = 0
A+B = -C
B X (A+B) = - B X C
BXA + 0 = - B X C
AXB = BXC ( ANS)
This answer is given by SARIKA and modified by admin only.
a small block slides down a smooth inclined plane , starting from rest at time t = 0 .let Sn be the distance travelled by the block in time interval t = n - 1 to t = n. then the ratio Sn / Sn + 1 is ??
distance travelled in n seconds = 1/2 an2 (a is the acceleration in the direction of incline)
distance travelled in n-1 second = 1/2 a(n-1)2
distance travelled in n+1 second = 1/2 a(n+1)2
so according to given condition Sn / Sn+1 = [1/2 an2 - 1/2 an2] / [1/2 a(n+1)2 - 1/2 an2]
now solve
1. A mass "m" moving with a velocity "u" hits a surface at an angle "Q" with the normal at the point of hitting . How much force does it exerts, if no energy is lost ?
2. A thin cicular loop of radius "R" rotates about its vertical diameter with angular frequency "W" . show that a small bead on the wire loop remains at its lowermost point for W ≤ (root of g/R) . what is the angle made by the radius vector joining the center to the bead with the vertical downwards direction for W= ( roots of 2g/R) . neglect friction .
3. A rear side of a truck is open and a box of 40 kg is placed 5m away from the open end . the coefficient of friction b/w box and surface in 0.15 on a straight road , the truck starts from rest and acclerates with 2m/s2. at what distance from the starting point does the box fall off the truck .
4. Straight from rest , a mass "m" slides down on inclined plane "Q" in a time "n times " the time to slide down the same lenght in absence of friction . Find the coefficient of friction .
1) angle is measured from vertical so vertical component of velocity = ucosθ
and since no energy is lost so returning velocity in vertical direction will be same
so impulse = (mucosθ) - (-mucosθ)
force cant be calculated without time of impact.
2)
at the position given in diagram mrω2cosθ = mgsinθ for equilibrium
so rω2cosθ = gsinθ and r = Rsinθ
so Rsinθω2cosθ = gsinθ so Rω2cosθ = g or ω = √[g/Rcosθ] so ω ≥ √[g/R]
thus for W ≤ (root of g/R), bead will remain at the lowermost point.
for second part compare W= ( roots of 2g/R) and ω = √[g/Rcosθ]
3) pseudo in opposite direction = ma
frictional force in the direction of motion of truck = 0.15mg
so equation of motion of box in opposite direction is ma - 0.15mg = ma'
so 2-1.5 = a' or a' = 0.5
distance = 5m
so time taken upto fall of box can be obtained by s = 1/2 a' t2
after calculating t, use again s = 1/2 (2) t2 for finding distance travelled by truck.
4) in first case with friction s = 1/2 (gsinθ-μgcosθ) (nt)2
in second case s = 1/2 (gsinθ) t2
now solve.
Projectile motions Plzz help!!!!?
2)If α = 30 degrees and β = 30 degrees and 'a'= 4.9 m, find the initial velocity of projection.
component of initial velocity along OB = ucos30
component of final velocity along OB = 0
component of g along OB = -gsin30
now use v = u + at
in second part clearly explain "a"
s1 = 0*10+1/2*a*100 = 50a
s1+s2 = 0*20+1/2*a*400 = 200a
now solve