The average acceleration vector for a particle having a uniform circular motion is
The average acceleration vector for a particle having a uniform circular motion is
- A
A constant vector of magnitude $\frac{{{v^2}}}{r}$
- B
A vector of magnitude $\frac{{{v^2}}}{r}$ directed normal to the plane of the given uniform circular motion
- C
Equal to the instantaneous acceleration vector at the start of the motion
- D
A null vector
Similar Questions
A particle is moving with constant speed $v$ in $x y$ plane as shown in figure. The magnitude of its angular velocity about point $O$ is .........
A particle is moving with constant speed $v$ in $x y$ plane as shown in figure. The magnitude of its angular velocity about point $O$ is .........
body moves with constant angular velocity on a circle. Magnitude of angular acceleration
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Can you associate vectors with $(a)$ the length of a wire bent into a loop, $(b)$ a plane area, $(c)$ a sphere ? Explain.
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A student skates up a ramp that makes an angle $30^{\circ}$ with the horizontal. $He /$ she starts (as shown in the figure) at the bottom of the ramp with speed $v_0$ and wants to turn around over a semicircular path xyz of radius $R$ during which he/she reaches a maximum height $h$ (at point y) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then ( $g$ is the acceleration due to gravity)
$(A)$ $v_0^2-2 g h=\frac{1}{2} g R$
$(B)$ $v_0^2-2 g h=\frac{\sqrt{3}}{2} g R$
$(C)$ the centripetal force required at points $x$ and $z$ is zero
$(D)$ the centripetal force required is maximum at points $x$ and $z$
A student skates up a ramp that makes an angle $30^{\circ}$ with the horizontal. $He /$ she starts (as shown in the figure) at the bottom of the ramp with speed $v_0$ and wants to turn around over a semicircular path xyz of radius $R$ during which he/she reaches a maximum height $h$ (at point y) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then ( $g$ is the acceleration due to gravity)
$(A)$ $v_0^2-2 g h=\frac{1}{2} g R$
$(B)$ $v_0^2-2 g h=\frac{\sqrt{3}}{2} g R$
$(C)$ the centripetal force required at points $x$ and $z$ is zero
$(D)$ the centripetal force required is maximum at points $x$ and $z$
- [IIT 2020]
Two bodies $A$ & $B$ rotate about an axis, such that angle $\theta_A$ (in radians) covered by first body is proportional to square of time, & $\theta_B$ (in radians) covered by second body varies linearly. At $t = 0, \theta \,A = \theta \,B = 0$. If $A$ completes its first revolution in $\sqrt \pi$ sec. & $B$ needs $4\pi \,sec$. to complete half revolution then; angular velocity $\omega_A : \omega_B$ at $t = 5\, sec$. are in the ratio
Two bodies $A$ & $B$ rotate about an axis, such that angle $\theta_A$ (in radians) covered by first body is proportional to square of time, & $\theta_B$ (in radians) covered by second body varies linearly. At $t = 0, \theta \,A = \theta \,B = 0$. If $A$ completes its first revolution in $\sqrt \pi$ sec. & $B$ needs $4\pi \,sec$. to complete half revolution then; angular velocity $\omega_A : \omega_B$ at $t = 5\, sec$. are in the ratio