Q: A boy is running over a circular track with uniform speed of 10 m/s. Find the average velocity for movement of boy from A to B( in m/s) − −

vav= Time Displacement =(πR/v)2R=π2v=π20ms−1

Q: The ratio of angular speeds of minute hand and hour hand of a watch is

60min→2π1 min→602π=30π12 h→2π1 h→122π=6π1 min→6×60π=12×30πNow ratio⇒30π×π12×30⇒12:1

Q: Q. 38. A point moves with deceleration along the circle of radius R so that at any moment of time its tangential and normal accelerations are equal in moduli. At the initial moment t = 0 the velocity of the point equals v0. Find: (a) the velocity of the point as a function of time and as a function of the distance covered s; (b) the total acceleration of the point as a function of velocity and the distance covered.

(a) According to the problem∣wt∣=∣wn∣ For v(t) , dt−dv=Rv2 Integrating this equation from v0≤v≤v and 0≤t≤t −∫v0vv2dv=R1∫0tdt or, v=(1+Rv0t)v0 Now for v(s)−dsvdv=Rv2 Integrating this equation from v0≤v≤v and 0≤s≤s So, ∫v0vvdv=−R1∫0sds or ,lnv0v=−Rs Hence v=v0e−s/R (b) The normal acceleration of the point wn=Rv2=Rv2e−2s/R (using 2)And as accordance with the problem ∣wt∣=∣wn∣ and wtu^t⊥wnu^n so, w=2wn=2Rv02e−2s/R=2Rv2

Q: Q. 39. A point moves along an arc of a circle of radius R. Its velocity depends on the distance covered s as v=as where a is a constant. Find the angle α between the vector of the total acceleration and the vector of velocity as a function of s .

From the equation, v=as wt=dtdv=2sadtds=2saas=2a2 and wn=Rv2=Ra2s As w , is a positive constant, the speed of the particle increases with time, and the tangential acceleration vector and velocity vector coincides in direction.Hence the angle between v and w is equal to between wtw^t an w, and α can be found by means of the formula tanα=∣wt∣∣wn∣=a2/2a2s/R=R2s

Q: The speed of a particle moving along a circular path of radius R=2 m varies with time as v=2t2 in SI units. The acceleration of particle at t=2 s is_____ m/s2.(Round off to the nearest integer)

ac=Rv2=2(2×22)2=32 m/s2 aτ=dtdv=4×2=8 m/s2 a=ac2+aτ2

Q: In the given figure, a=15 m/s2 represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R=2.5 m at a given instant of time. The speed of the particle is

acos30∘=rv2→1523=2.5v2→v=5.7 m/s

Q: A particle is moving on a circular path with constant speed v. It moves between two points A and B, which subtends an angle 60∘ at the centre of circle. The magnitude of change in its velocity and change in magnitude of its velocity during motion from A to B are respectively

Δv=2vsin2θ =2v×sin(260) ∣Δv∣=v Change in magnitude of velocity =0

Question 1

A boy is running over a circular track with uniform speed of 10 m/s. Find the average velocity for movement of boy from A to B( in m/s)
−

−

Accepted Answer

vav​= Time  Displacement ​=(πR/v)2R​=π2v​=π20​ms−1

Question 2

The ratio of angular speeds of minute hand and hour hand of a watch is

Accepted Answer

60min→2π1 min→602π​=30π​12 h→2π1 h→122π​=6π​1 min→6×60π​=12×30π​Now ratio⇒30π​×π12×30​⇒12:1

Question 3

A particle is moving around a circular path with uniform angular speed (ω). The radius of the circular path is (r). The acceleration of the particle is :-

Accepted Answer

The particle has only centripetal acceleration and no tangential acceleration since it is moving with a uniform speed ω r around the circle .Therefore, acceleration of the particle =ω2r=ω(ωr)=ωV

Question 4

Q. 38. A point moves with deceleration along the circle of radius R so that at any moment of time its tangential and normal accelerations are equal in moduli. At the initial moment t = 0 the velocity of the point equals v0. Find: (a) the velocity of the point as a function of time and as a function of the distance covered s; (b) the total acceleration of the point as a function of velocity and the distance covered.

Accepted Answer

(a) According to the problem∣wt​∣=∣wn​∣ For v(t) ,  dt−dv​=Rv2​ Integrating this equation from  v0​≤v≤v and 0≤t≤t −∫v0​v​v2dv​=R1​∫0t​dt or, v=(1+Rv0​t​)v0​​ Now for v(s)−dsvdv​=Rv2​  Integrating this equation from v0​≤v≤v and 0≤s≤s So,   ∫v0​v​vdv​=−R1​∫0s​ds or ,lnv0​v​=−Rs​ Hence  v=v0​e−s/R (b) The normal acceleration of the point wn​=Rv2​=Rv2e−2s/R​ (using 2)And as accordance with the problem  ∣wt​∣=∣wn​∣ and wt​u^t​⊥wn​u^n​ so, w=2​wn​=2​Rv02​​e−2s/R=2​Rv2​

Question 5

Q. 39. A point moves along an arc of a circle of radius R. Its velocity depends on the distance covered  s as v=as​ where a is a constant. Find the angle α between the vector of the total acceleration and the vector of velocity as a function of s .

Accepted Answer

From the equation, v=as​ wt​=dtdv​=2s​a​dtds​=2s​a​as​=2a2​ and wn​=Rv2​=Ra2s​ As w , is a positive constant, the speed of the particle increases with time, and the tangential acceleration vector and velocity vector coincides in direction.Hence the angle between v and w is equal to between wt​w^t​ an w, and α can be found by means of the formula  tanα=∣wt​∣∣wn​∣​=a2/2a2s/R​=R2s​

Question 6

Q. 41. A point moves in the plane so that its tangential acceleration wτ = a, and its normal acceleration wn  = bt^4, where a and b are positive constants, and t is time. At the moment t = 0 the point was at rest. Find how the curvature radius R of the point's trajectory and the total acceleration w depend on the distance covered s.

Accepted Answer

As wt =a and at t=0 , the point is at restSo, v(t) and s(t) are, v=at and s=21​at2   …  (1)Let R be the curvature radius, then wn​=Rv2​=Ra2t2​=R2as​ (using 1)But according to the problemwn​=bt4  So, bt4=Ra2t2​ or, R=bt2a2​=2bsa2​ (using 1) Therefore, w=wt2​+wn2​​=a2+(2as/R)2​=a2+(4bs2/a2)2​ (using 2)   Hence, w=a1+(4bs2/a3)2​

Question 7

Q. 42. A particle moves along the plane trajectory y(x) with velocity v whose modulus is constant. Find the acceleration of the particle at the point x=0 and the curvature radius of the trajectory at that point if the trajectory has the form (a) of a parabola y=ax2 ; (b) of an ellipse (x/a)2+(y/b)2=1 ; a and b are constants here.

Accepted Answer

(a) Let us differentiate twice the path equation y (x) with respect to time.dtdy​=2axdtdx​;dt2d2y​=2a[(dtdx​)2+xdt2d2x​] Since the particle moves uniformly, its acceleration at all points of the path is normal and at the point  x=0  it coincides with the direction of derivative d2y/dt2.  Keeping in mind that at the point  x=0,∣∣​dtdx​∣∣​=v1​ We get  x=0,∣∣​dtdx​∣∣​=v1​ So, wn​=2av2=Rv2​, or R=2a1​ Note that we can also calculate it from the formula of problem (1.35 b)(b) Differentiating the equation of the trajectory with respect to time we see thatb2xdtdx​+a2ydtdy​=0 …   (1)which implies that the vector (b2xi+a2yJ)  is normal to the velocity vector  v=dtdx​i+dtdy​j​ which, of course, is along the tangent. Thus the former vector is along the normal and the normal component of acceleration is clearlywn​=(b4x2+a4y2)1/2b2xdt2d2x​+a2ydt2d2y​​ on using wn​=w⋅n/∣n∣ At x=0,y=±b and so at x=0 wn​=±dt2d2y​∣∣​x=0​ Differentiating (1)b2(dtdx​)2+b2x(dt2d2x​)+a2(dtdy​)2+a2y(dt2d2y​)=0 Also from (1) dtdy​=0 at x=0 So,   (dtdx​)=±v  (since tangential velocity is constant = v)Thus (dt2d2y​)=±a2b​v2 and  ∣wn​∣=a2bv2​=Rv2​ This gives R=a2/b .

Question 8

In the given figure, particle is travelling anticlockwise in a circle of radius 10 m. The acceleration vector is indicated at certain instant. The magnitude of velocity v of particle at this instant is______ m/s.

Accepted Answer

10v2​=50cos37∘
v=20 m/s

Question 9

The speed of a particle moving along a circular path of radius R=2 m varies with time as v=2t2 in SI units. The acceleration of particle at t=2 s is_____ m/s2.(Round off to the nearest integer)

Accepted Answer

ac​=Rv2​=2(2×22)2​=32 m/s2   aτ​=dtdv​=4×2=8 m/s2  a=ac2​+aτ2​​

Question 10

In the given figure, a=15 m/s2 represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R=2.5 m at a given instant of time. The speed of the particle is

Accepted Answer

acos30∘=rv2​→1523​​=2.5v2​→v=5.7 m/s

Question 11

A particle is moving on a circular path with constant speed v. It moves between two points A and B, which subtends an angle 60∘ at the centre of circle. The magnitude of change in its velocity and change in magnitude of its velocity during motion from A to B are respectively

Accepted Answer

Δv=2vsin2θ​
=2v×sin(260​)
∣Δv∣=v
Change in magnitude of velocity =0

Question 12

A particle of mass m moves in a circular orbit under the central potential field, U(r)=C/r , where C is a positive constant. The correct - velocity graph of the particle's motion is :

Accepted Answer

​F(x)=−∂x∂u​=x2C​v=xmC​​;v∝x1/21​​

Question 13

Assertion: In circular motion work done by all the forces acting on the body is zero. Reason: Centripetal force and velocity are mutually perpendicular.

Accepted Answer

In non-uniform circular motion (when speed = constant) work done by all the forces is not zero.

Question 14

A particle moves in a circle of radius 0.5 m at a speed that uniformly increases. Find the angular acceleration of particle if its speed changes from 2.0 m/s to 4.0 m/s in 4.0 s.

Accepted Answer

The tangential acceleration of the particle isat​=dtdv​=4.04.0−2.0​=0.5 m/s2The angular acceleration isα=rat​​=0.50.5​=1rad/s2

Question 15

The speed of a particle moving in a circle of radius r=2 m varies with time t as v=t2, where t is in second and v in m/s. Find the radial, tangential and net acceleration at t=2 s.

Accepted Answer

Linear speed of particle at t=2 s isv=(2)2=4 m/s∴ Radial accelerationar​=rv2​=2(4)2​=8 m/s2The tangential acceleration is at​=dtdv​=2t∴ Tangential acceleration at t=2 s isat​=(2)(2)=4 m/s2∴ Net acceleration of particle at t=2 s isa=(ar​)2+(at​)2​=(8)2+(4)2​ or a=80​ m/s2

Question 16

In circular motion, what are the possible values (zero, positive or negative) of the following :(a) ω⋅v(b) v⋅a(c) ω⋅α

Accepted Answer

(a) v lies in the plane of circle and ω is always perpendicular to this plane.∴ v⊥ω (always)Hence, ω⋅v is always zero.(b) v and a both lie in the plane of circle and the angle between these two vectors may be acute (when speed is increasing) obtuse (when speed is decreasing) or 90∘ (when speed is constant).Hence, v⋅a may be positive, negative or zero.(c) ω and α are either parallel (θ=0∘ between ω and α) or antiparallel (θ=180∘). In uniform circular motion, α has zero magnitude. Hence, ω⋅α may be positive, negative or zero.

Filters

Filters

class

subject

topic

concept

difficulty

Solution for Class 11, Physics, Motion in a Plane

Found the solution, but did not understand the concept?

Ask my first question