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Chapter 3: Problem 83

A particle is moving with a constant angular acceleration of \(4 \mathrm{rad} / \mathrm{s}^{2}\) in a circular path. At \(t=0\), particle was at rest. Find the time at which the magnitudes of centripetal acceleration and tangential acceleration are equal. \(\begin{array}{llll}\text { (A) } 1 \mathrm{~s} & \text { (B) } 2 \mathrm{~s} & \text { (C) } \frac{1}{2} \mathrm{~s} & \text { (D) } \frac{1}{4} \mathrm{~s}\end{array}\)

Short Answer

Expert verified
The magnitudes of centripetal acceleration and tangential acceleration are equal at t = 1 s. The correct answer is \( \boxed{\text{(A) } 1 \mathrm{~s}} \).

Step by step solution

01

Write the given information and required formulas

We are given: - 伪 = 4 rad/s虏 - a_t = 伪r - 蠅 = 伪t (initial angular velocity is 0) - a_c = 蠅虏r We want to find the time t when a_t = a_c.
02

Equate tangential and centripetal accelerations

Equate a_t and a_c and substitute the values: 伪r = 蠅虏r Since 伪 = 4 rad/s虏 and 蠅 = 伪t, the equation becomes: 4r = (4t)虏r
03

Solve for t

Solve the above equation for t: 4r = 16t虏r Divide both sides by 4r: t虏 = 1 Take the square root of both sides: t = 卤1 s Since time can only be positive, we discard the negative value: t = 1 s
04

Conclusion

The magnitudes of centripetal acceleration and tangential acceleration are equal at t = 1 s. The correct answer is (A) 1 s.

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Most popular questions from this chapter

A block of mass \(0.1 \mathrm{~kg}\) is held against a wall by applying a horizontal force of \(5 \mathrm{~N}\) on the block. If the co-efficient of friction between the block and the wall is \(0.5\), the magnitude of the frictional force acting on the block is (A) \(2.5 \mathrm{~N}\) (B) \(0.98 \mathrm{~N}\) (C) \(4.9 \mathrm{~N}\) (D) \(0.49 \mathrm{~N}\)

A body of mass \(1.5 \mathrm{~kg}\) is thrown vertically upwards with an initial velocity of \(40 \mathrm{~m} / \mathrm{s}\) reaches its highest point after \(3 \mathrm{~s}\). The air resistance acting on the body during the ascent is (assuming air resistance to be uniform, \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(35 \mathrm{~N}\) (B) \(25 \mathrm{~N}\) (C) \(15 \mathrm{~N}\) (D) \(5 \mathrm{~N}\)

A pebble of mass \(0.05 \mathrm{~kg}\) is thrown vertically upwards. The direction and magnitude of the net force on the pebble is given below, choose the incorrect option. (A) During its upward motion, force is \(0.5 \mathrm{~N}\) in vertically upward. (B) During its downward motion, force is \(0.5 \mathrm{~N}\) in vertically downward. (C) At the highest point, where it is momentarily at rest, force is \(0.5 \mathrm{~N}\) in vertically downward. (D) If the pebble was thrown at an angle of say \(45^{\circ}\) with the horizontal direction, force is \(0.5 \mathrm{~N}\) in vertically downward (Ignoring air resistance).

A block of mass \(M\) is pulled along a horizontal frictionless surface by a rope of mass \(m\). If a force \(P\) is applied at the free end of the rope, the force exerted by the rope on the block is (A) \(\frac{P m}{M+m}\) (B) \(\frac{P m}{M-m}\) (C) \(P\) (D) \(\frac{P M}{M+m}\)

A constant retarding force of \(50 \mathrm{~N}\) is applied to a body of mass \(20 \mathrm{~kg}\) moving initially with a speed of \(15 \mathrm{~ms}^{-1}\). How long does the body take to stop? (A) \(2 \mathrm{~s}\) (B) \(4 \mathrm{~s}\) (C) \(6 \mathrm{~s}\) (D) \(8 \mathrm{~s}\)
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