Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A sphere of mass 20 kg is suspended by a metal wire of unstretched length 4 m and diameter 1 mm. When in equilibrium, there is a clear gap of 2 mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle θ with the vertical and is released. Find the maximum value of θ so that the sphere does not rub the floor. Young modulus of the metal of the wire is 2.0 × 1011 N m-2. Make appropriate approximations.

A sphere of mass 20 kg is suspended by a metal wire of unstretched length 4 m and diameter 1 mm. When in equilibrium, there is a clear gap of 2 mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle θ with the vertical and is released. Find the maximum value of θ so that the sphere does not rub the floor. Young modulus of the metal of the wire is 2.0 × 1011 N m-2. Make appropriate approximations.

Grade:upto college level

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
6 years ago
Sol. . At equilibrium ⇒ T = mg When it moves to an angle θ, and released, the tension the T’ at lowest point is ⇒ T’ = mg + mv2/r The change in tension is due to centrifugal force ∆T = mv2/r ….(1) ⇒ Again, by work energy principle, ⇒ 1/2mv2 – 0 = mgr(1 - cosθ) ⇒ v2 = 2gr (1- cosθ) So, ∆T = m[2gr(1 – cos θ)]/r = 2mg(1 – cos θ) ⇒ F = ∆T ⇒ F = YA ∆L/L = 2mg – 2mg cos θ ⇒ 2mg cos θ = 2mg - YA ∆L/L = cos θ = 1 - YA ∆L/L(2mg)

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free