Answer: Option C. -> constant kinetic energy : C We can see that direction of velocity and acceleration changes continuously ∴ Both are not constant. Momentum is m.⃗v which depends on velocity and changes with velocity. The only quantity that is constant is speed and kinetic energy is 12m|⃗v|2 where|⃗v| is the speed. ∴ Answer is kinetic energy.
Question 2. At t1 = 2.00 s, the acceleration of a particle in counter clockwise circular motion is It moves at constant speed. At time t2 = 5.00 s, the particle's acceleration is What is the radius of the path taken by the particle if t2 - t1 is less than one period?
Answer: Option A. -> 40π2 : A Att1=2s,→a=(5√3ms2)^i+5^j Att2=5s,→a=(5ms2)^i−(5√3ms2)^i+^j Given the body is moving in anti-clockwise direction Since,the speed is constant, net acceleration will always be directed toward the centre and since it is This will be the only path possible using this, we can calculate ω as we can find thechange in angle tanθ1=15√3=13 tanθ2=3√35=√3 ∴θ1=30∘ ⇒θ2=60∘ ∴ We can see that angle between the vectors is 90∘ which means particle would have covered an angle of 360−90=270∘=3π/2radin3s ∴ω=ΔθΔt=3π23=π2rad/s a=ω2r r=aω2=√(5√3)2+52(π2)2=√100π24 =4×10π2=40π2
Question 3. Two cars having masses m1andm2 move in circles of radii r1andr2 respectively. If they complete the circle in equal time, the ratio of their angular speeds ω1ω2 is _______
Answer: Option D. -> 1 : D ω1=△θ△t If time taken to complete a revolution is t. Then ω1=2πt,ω2=2πt ⇒t=2πω1=2πω2 ⇒ we get ω1=ω2orω1ω2=1
Question 4. A circular disc is rotating with constant angular velocity ω about an axis that passes through the centre. A particle 'P' is kept at a distance of 2m from the centre and another particle 'Q' at a distance of 3m from the centre. Which of these will have higher centripetal acceleration?
Answer: Option B. -> Q : B If a particle goes in a circle with constant ω then it has a centripetal acceleration given by ω2R since ω is constant for both particles. So the acceleration depends on R ∵RQ>RP ⇒ω2RQ>ω2RP So particle q will have higher centripetal acceleration.
Question 5. A car is moving on a circular path and takes a turn. If R1andR2 be the reactions on the inner and outer wheels respectively, then
Answer: Option B. -> R1 : B Reaction on inner wheelR1=12M[g−v2hra] Reaction on outer wheel R2=12M[g+v2hra] where, r = radius of circular path, 2a = distance between two wheels and h = height of centre of gravity of car.
Question 6.A stone of mass mis tied to a string of length land rotated in a circle with a constant speed v. If the string is released, the stone flies
Answer: Option C. -> Tangentially outward : C Stone flies in the direction of instantaneous velocity due to inertia
Question 7. A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2)^i+ (4.00 m/s2)^j. At that instant and in unit-vector notation, what is the acceleration of the wallet? IIT JEE- 2001
Answer: Option C. -> 3 ^i + 6 ^j : C dθdt is constant. In other words in uniform circular motion the angular velocity remains constant body doesn't have any tangential acceleration but normal acceleartion. aN=v2Rorω2R ForpurseaN=√(2)2+(4)2=√20;R=2 ⇒√20=ω22 ⇒ω2=√5 ForwalletaN=ω2R Hence ωis same But~R=3 ⇒aN=√5×3 aN=3√5 So the above answer matches with the magnitude of third option in the given answers.
Question 8.A car is travelling with speed v on a circular road of radius r. If it is increasing its speed at the rate of 'a' meter/sec2, then the resultant acceleration will be
Answer: Option B. -> 20 π m / minute : B The linear speed is, v = r ω ;ω = θt Substituting, we get, v=20 π m / minute
Question 10. A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 5.00 s, it is at point (5.00 m, 6.00 m) with velocity (3π m/s)^jand acceleration in the positive x direction. At time t2 = 10.0 s, it has velocity (-3π m/s)^i and acceleration in the positive y direction. What are the (a) x and (b) y coordinates of the center of the circular path if t2 - t1 is less than one period?
Answer: Option A. -> (15 m, 6 m) : A We know the body is undergoing uniform circular motion, ∴ only possible path is As acceleration should always be directed toward the centre To calculate the position of the centre, we need the radius, to calculate radius,v=ωr we can use the relation,as we can calculate ω Δθ=3π2,Δt=5s =3π2×5=3π10S−1 usingv=ωr r=vω |→v|=3πm/sgiven ∴r=3π3π10=30π3π=10m ∴ifr=10m, coordinates of the centre (15 m,6m)