Name: Collision Between a Point Mass and a Rigid Body
Uploaded: 2019-07-19
Duration: 7 min 57 s
Description: Can we treat the collision of a point mass with a rigid body in the same way we treated the collision of point masses? Watch this video to know about it.

Question

A point $A$ is located on the rim of a wheel of radius $R = 0.50 m$ which rolls without slipping along a horizontal surface with velocity $v = 1.00 m / s$ as shown in figure. Find:

(a) the modulus and the direction of the acceleration vector of the point $A$ ;

(b) the total distance s traversed by the point $A$ between the two successive moments at which it touches the surface.

Accepted Answer

(a). At point A the instantaneous velocity of point of contact is zero and acceleration is also zero. but it is in the rotation mode due to its angular velocity, therefore acceleration will be in vertical direction which is due to centripetal force directed towards centre.

$\vec{a} = R ω^{2} \hat{j} v = R ω \Rightarrow ω = \frac{1}{0.5} = 2 \Rightarrow \vec{a} = 0.5 \times 2^{2} \hat{j} \Rightarrow \vec{a} = 2 \hat{j} {ms}^{- 2}$

(b).

Let us assume, at any point on the rim of the wheel making an angle $θ$ with the vertical

Magnitude of the velocity of the point P $v_{P} = {{[v_{c m}}^{2} + {(R ω)}^{2} + 2 v_{c m} R ω \cos (θ)]}^{\frac{1}{2}}$

In case of rolling $R = 0.50 m$

${v^{2}}_{P} = v^{2} + v^{2} + 2 v^{2} \cos (θ) \Rightarrow {v^{2}}_{P} = 2 v^{2} (1 + \cos (θ) \Rightarrow v_{P} = 2 v \cos (\frac{θ}{2})$

Now for displacement $d s = v_{P} d t \Rightarrow d s = 2 v \cos (\frac{θ}{2}) d t \Rightarrow d s = 2 v \cos (\frac{ω t}{2}) d t θ = ω t \Rightarrow s = 2 v \int_{0}^{\frac{2 π}{ω}} \cos (\frac{ω t}{2}) d t \Rightarrow s = \frac{4 v}{ω} {|\sin (\frac{ω t}{2})|}_{0}^{\frac{2 π}{ω}} v = R ω s = 8 R$

Important Questions on Rotational Mechanics

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