Curvilinear Motion MCQ Quiz in मल्याळम - Objective Question with Answer for Curvilinear Motion - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Tangential acceleration = r × angular acceleration

Now,

Now, Normal acceleration

∴ Resultant acceleration

Explanation:

Explanation:

Radial and Transverse co-ordinates:

• In this system, the position of the particle is defined by the polar coordinates r and θ
• Further, the velocity and acceleration of the particle are resolved into components along and perpendicular to the position vector . These components are called:

1. Radial components denoted by 
2. Transverse components denoted by 

The unit vectors of  and  are represented by  and  respectively. Here A₁A₂ and B₁B₂ are perpendicular to the x-axis and are drawn from the endpoints A₁ and B₁ of the unit vectors  and  respectivel

 
Furthur: PA₁ =  and PB₂ = 
That gives: A₁A₂ = Sin θ and B₁B₂ = Cos θ, PA₂ = Cos θ and PB₂ = Sin θ
Applying the triangle law of addition of vectors

When the above identities are differentiated with respect to time t.

Denoting , we can write 

In terms of the position vector , the velocity vector is defined as

Replacing  as worked out above,

, 
Transverse component, 

Acceleration:

Substituting the values for  and  in acceleration.

CONCEPT:

Centripetal Acceleration (a_c): 

• Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration.
• It always acts on the object along the radius towards the center of the circular path.
• The magnitude of centripetal acceleration,

Where v = velocity of the object and r = radius

Tangential acceleration (a_t):

• It acts along the tangent to the circular path in the plane of the circular path.
• Mathematically Tangential acceleration is written as

Where α = angular acceleration and r = radius

CALCULATION:

Given – v = 10 m/s, r = 25 m and a_t = 3 m/s²

• Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration i.e.,

Centripetal Acceleration (a_c):

Hence, net acceleration 

Since the wheel is in constant contact with the ground, the length of the arc correlating to the angle is also equal to x.

Rolling without slipping condition: x = Rθ.

Time Period

Equation of path,

θ = 60°, x = 35m, y = 8 -2 = 6m

⇒ u₀ = 20.97 m/s

U_0x = u₀cosθ& u_0y = u₀sinθ

X = u₀ t = (u₀cosθ )t

 -----------(1)

Y = u_0y t – ½ gt² --------- (2)

Substitute equation (1) is equation (2)

Explanation:

Radius of Curvature:
We can draw a circle that closely fits nearby points on a local section of a curve, as follows

• We say the curve and the circle osculate since the 2 curves have the same tangent and curvature at the point where they meet.
• The radius of curvature of the curve at a particular point is defined as the radius of the approximating circle. This radius changes as we move along the curve.
• The formula for the radius of curvature at any point x for the curve y = f(x) is given by:

​Radius of curvature = 
So, option (3) is the correct answer.

Tangential component of acceleration is responsible for change in magnitude of velocity. If magnitude of velocity is constant then a_t = O.